Strength of materials_by_r_s_khurmi-601-700

Chapter 25 : Theorem of Three Moments 601
Since iB is equal to –i′B, therefore E · iB is equal to – E · i′B.
or 1 1 1
1 1 1
( 2 )
6A B
a x l
M M
I l I
+ + = 2 2 2
2 2 2
( 2 )
6C B
a x l
M M
I l I
⎡ ⎤
− + +⎢ ⎥
⎣ ⎦
(MA + 2MB)
1
1
l
I + (MC+2MB) 2
2
l
I
= – 1 1 2 2
1 1 2 2
6 6a x a x
I l I l
−
1 1 2 2
1 1 2 2
2 2A B C B
l l l l
M M M M
I I I I
+ + + = – 1 1 2 2
1 1 2 2
6 6⎛ ⎞
+⎜ ⎟⎝ ⎠
a x a x
I l I l
∴ 1 1 2 2
2 1 2 2
2A B C
l l l l
M M M
I I I I
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=
1 1 2 2
1 1 2 2
6 6a x a x
I l I l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
NOTES: 1. For the sake of simplicity, we have considered a continuous beam with two spans only. But this
equation can be extended for any number of spans.
2. If moment of inertia of the beam is constant, then
MA l1 + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 2
6 6a x a x
I I
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
3. The shear force diagram for the beam may be drawn as usual.
4. The elastic curve of the beam may be drawn as usual as shown in Fig. 25.1 (c).
25.4. Application of Clapeyron’s Theorem of Three Moments to
Various Types of Continuous Beams
We have already studied in Art. 25.3 the Clapeyron’s theorem of three moments. Now we shall
discuss its application to the following types of continuous beams:
1. Continuous beams with simply supported ends,
2. Continuous beams with fixed end supports,
3. Continuous beams with the end span overhanging and
4. Continuous beams with a sinking support.
25.5. Continuous Beams with Simply Supported Ends
Sometimes, a continuous beam is simply supported on its one or both the end supports. In such a
case, the fixing moment on the simply supported end is zero.
EXAMPLE 25.1. A continuous beam ABC 10 m long rests on three supports A, B and C at the
same level and is loaded as shown in Fig. 25.2.
Fig. 25.2
Determine the moments over the beam and draw the bending moment diagram. Also calculate
the reactions at the supports and draw the shear force diagram.
SOLUTION. Given : Length AB (l1) = 6 m ; Length BC (l2) = 4 m ; Point load at D = (W) = 3 kN;
Distance AD (a) = 2 m ; Distance DB (b) = 4 m and uniformly distributed load in BC = w = 1 kN/m.
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602 Strength of Materials
Fig. 25.3
Moments over the beam
Let *MA = Fixing moment at A,
MB = Fixing moment at B and
*MC = Fixing moment at C.
First of all, let us consider the beam AB as a simply supported. Therefore, bending moment at D,
MD =
1
3 2 4
4
6
Wab
l
× ×
= = kN-m
Similarly, bending moment at the mid of the span BC
=
2 2
2 1 (4)
2
8 8
wl ×
= = kN-m
Now draw the μ-diagrams with the help of above bending moments as shown in Fig. 25.3 (a).
From the geometry of the above bending moment diagrams, we find that
1 1a x =
2 21 1 4
2 4 4 4 2 32
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × + × × + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
and 2 2a x =
2 32
2 4 2
3 3
⎛ ⎞× × × =
⎝ ⎠
Now using three moments equation,
MA l1 + 2MB (l1 + l2) + MC l2 = 1 1 2 2
1 2
6 6a x a x
l l
⎛ ⎞
+⎜ ⎟
⎝ ⎠
0 + 2 MB (6 + 4) + 0 =
32
66 32 3
6 4
⎛ ⎞××⎜ ⎟+
⎜ ⎟
⎜ ⎟⎝ ⎠
* Since the beam is simply supported at A and C, therefore fixing moments MA and MC will be zero.
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∴ 20 MB = – (32 + 16) = – 48
or MB =
48
20
− = – 2.4 kN-m Ans.
Now complete the bending moment diagram as shown in Fig. 25.3 (b).
Shear force diagram
Let RA = Reaction at A,
RB = Reaction at B and
RC = Reaction at C.
Taking moments about B,
RA × 6 – (3 × 4) = – 2.4 ...(ä MB = – 2.4 kN-m)
∴ RA =
2.4 12.0 9.6
6 6
− +
= = 1.6 kN Ans.
Similarly, RC × 4 – (4 × 2) = – 2.4 ...(ä MB = – 2.4 kN-m)
∴ RC =
2.4 8.0 5.6
4 4
− +
= = 1.4 kN Ans.
and RB = (3 + 1 × 4) – (1.6 + 1.4) = 4.0 kN Ans.
Now complete the shear force diagram as shown in Fig. 25.3 (c)
EXAMPLE 25.2. A continuous beam ABCD, simply supported at A, B, C and D, is loaded as
shown in Fig. 25.4.
Fig. 25.4
Find the moments over the beam and draw bending moment and shear force diagrams.
SOLUTION. Given : Length AB (l1) = 6 m ; Length BC (l2) = 5 m ; Length CD (l3) = 4 m ; Load
at E (W1) = 9 kN ; distance AE (a1) = 2 m ; Distance EB (b1) = 4 m ; Load at F (W2) = 8 kN ; Distance
BF (a2) = 2 m ; Distance FC (b2) = 3 m and uniformly distributed load in CD (w) = 3 kN/m.
Moments over the beam
Let * MA = Fixing moment at A,
MB = Fixing moment at B,
MC = Fixing moment at C and
*MD = Fixing moment at D.
First of all, let us consider the beam AB as a simply supported beam. Therefore bending moment
at E,
ME =
1 1 1
1
9 2 4
6
W a b
l
× ×
= = 12 kN-m
Similarly, MF = 2 2 2
2
8 2 3
5
W a b
l
× ×
= = 9.6 kN-m
* Since the beam is simply supported at A and D, therefore fixing moments MA and MD will be zero.
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and bending moment at the mid of the span CD
=
2 2
3 3 (4)
8 8
wl ×
= = 6 kN-m
Fig. 25.5
Now draw the m-diagrams with the help of above bending moments as shown in Fig. 25.5 (b).
From the geometry of the above bending moment diagrams, we find that for the spans AB and BC,
1 1a x =
2 21 1 4
2 12 4 12 2 96
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞× × × + × × + =⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦
and 2 2a x =
2 31 1 2
3 9.6 2 9.6 3 64
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞× × × + × × + =⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦
Similarly, for the spans BC and CD,
* 2 2a x =
2 31 1 3
3 9.6 2 9.6 2 56
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞× × × + × × + =⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦
and 3 3a x =
2
6 4 2 32
3
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
Now using three moments equation for the spans AB and BC,
MA l1 + 2MB (l1 + l2) + MC l2 = 1 1 2 2
1 2
6 6a x a x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
0 + 2 MB (6 + 5) + MC × 5 =
6 96 6 64
6 5
× ×⎛ ⎞
− +⎜ ⎟
⎝ ⎠
22 MB + 5 MC = – 172.8 ...(i)
* The previous value of 2 2a x is with reference to the support C (being the end support of spans AB and
BC). this value of 2 2a x is with reference to the support B (being the end support of spans BC and CD).
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Again using three moments equation for the spans BC and CD,
MB l2 + 2 MC (l2 + l3) + MD l3 =
3 32 2
2 3
66 a xa x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
MB × 5 + 2 MC (5 + 4) + 0 =
6 56 6 32
5 4
× ×⎛ ⎞
+⎜ ⎟
⎝ ⎠
5 MB + 18 MC = – 115.2 ...(ii)
Solving equations (i) and (ii),
MB = – 6.84 kN-m and MC = – 4.48 kN-m Ans.
Reactions at the supports
RC = Reaction at B,
RC = Reaction at C and
RD = Reaction at D.
RA × 6 – (9 × 4) = – 6.84 ...(ä MB = – 6.84 kN-m)
∴ RA =
6.84 36 29.16
6 6
− +
= = 4.86 kN Ans.
Now taking moments about C,
RD × 4 – (12 × 2) = – 4.48 ...(ä MC = – 4.48 kN-m)
∴ RD =
4.48 24 19.52
4 4
− +
= = 4.88 kN Ans.
Again taking moments about C,
RA × 11 – (9 × 9) + RB × 5 – (8 × 3)
= – 4.48 ...(ä MC = – 4.48 kN-m)
4.86 × 11 – 81 + 5 RB – 24= – 4.48
RB =
4.48 53.46 81 24 47.06
5 5
− − + +
= = 9.41 kN Ans.
and RC = (9 + 8 + 12) – (4.86 + 4.88 + 9.41) = 9.85 kN Ans.
Now draw the shear force diagram as shown in Fig. 25.5 (c)
25.6. Continuous Beams with Fixed End Supports
Sometimes, a continuous beam is fixed at its one or both ends. If the beam is fixed at the left end
A, then an imaginary zero span is taken to the left of A and the three moments theorem is applied as
usual. Similarly, if the beam is fixed at the right end, then an imaginary zero span is taken after the
right end support and the three moments theorem is applied as usual.
NOTES. 1. The fixing moment, at the imaginary support of the zero span i.e., M0 is always equal to zero.
2. Propped cantilevers and beams may also be analyses by Clapeyron’s theorem of three moments.
EXAMPLE 25.3. A continuous beam ABC of uniform section, with span AB as 8 m and BC as
6 m, is fixed at A and simply supported at B and C. The beam is carrying a uniformly distributed
load of 1 kN/m throughout its length. Find the moments along the beam and the reactions at the
supports. Also draw the bending moment and shear force diagrams.
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SOLUTION. Given : Length AB (l1) = 8 m ; Length BC (l2) = 6 m and uniformly distributed load
(w) = 1 kN/m.
Moments along the beam
Since the beam is fixed at A, therefore assume a zero span to the left of A.
Let * M0 = Fixing moment at the left hand support of zero span,
MA = Fixing moment at A,
*MC = Fixing moment at C.
First of all, consider the beam AB as a simply supported beam. Therefore bending moment at the
mid of the span AB.
=
2 2
1 10 (8)
8 8
wl +
= = 8.0 kN-m
=
2 2
2 1 (6)
8 8
wl ×
= = 4.5 kN-m
Now draw the μ-diagram with the help of above bending moments as shown in Fig. 25.6 (b).
From the geometry of the above bending moment diagrams, we find that for the span 0A and AB,
Fig. 25.6
0 0a x = 0
and 1 1a x =
2 512
8 8 4 170.67
3 3
⎛ ⎞× × × = =⎜ ⎟
⎝ ⎠
* Since threre is a zero span on the left of A, therefore the fixing moment M0 will be zero. Moreover, as the
beam is simply supported at C, therefore fixing moment MC will also be zero.
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Similarly, for the span AB and BC,
1 1a x =
2 512
8 8 4 170.67
3 3
⎛ ⎞× × × = =⎜ ⎟
⎝ ⎠
and 2 2a x =
2
4.5 6 3 54
3
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
Now using three moments equation for the spans 0A and AB,
M0 l0 + 2 MA (0 + l1) + MB l1 =
0 0 1 1
0 1
6 6a x a x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
0 + 2 MA (0 + 8) + MB × 8 =
6 170.67
0
8
×⎛ ⎞
− +⎜ ⎟
⎝ ⎠
16 MA + 8 MB = – 128 ...(i)
Again using three moments equation for the spans AB and BC,
MA ll + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 2
6 6a x a x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
MA × 8 + 2 MB (8 + 6) + 0 =
6 170.67 6 54
8 6
× ×⎛ ⎞
− +⎜ ⎟
⎝ ⎠
8 MA + 28 MB = – 182 ...(ii)
Solving the equations (i) and (ii)
MA = – 5.75 kN-m and MB = – 4.5 kN-m Ans.
Reactions at supports
RB = Reaction at B, and
RC = Reaction at C,
RC × 6 – (6 × 3) = – 4.5 (ä MB = – 4.5 kN-m)
∴ RC =
4.5 18 13.5
6 6
− +
= = 2.2 kN Ans.
Now taking moments about A,
RC × 14 + RB × 8 – (1 × 14 × 7) =– 5.75 (ä MA = – 5.57 kN-m)
(2.25 × 14) – 8 · RB – 98 = – 5.75
∴ RB =
5.75 98 31.5
8
− + −
= 7.6 kN Ans.
and *RA = 14.0 – (2.25 + 7.6) = 4.15 kn Ans.
Now completed the shear force diagram as shown in Fig. 25.7 (c).
* The reaction at RA may also be found out by taking mokments about B, i.e.,
RA × 8 – (8 × 4) – MA = – 4.5 ...(ä MB = – 4.5 kN-m)
8 RA – 32 – 5.75 = – 4.5
∴ RA =
4.5 32 5.75 33.25
8 8
− + +
= = 4.15 kN Ans.
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EXAMPLE 25.4. Evaluate the bending moment and shear force diagrams of the beam shown
in Fig. 25.7.
Fig. 25.7
What are the reactions at the supports?
SOLUTION. Given : Length AB (l1) = 6 m ; Length BC (l2) = 6 m ; Uniformly distributed load in
AB (w) = 2 kN/m and point load at D (W) = 12 kN.
Bending moment at A, B and C
Since the beam is fixed at A and C, therefore assume a zero span to the left of A and right of C.
Let M0 = Fixing moments at the imaginary supports of zero span (on
the left of A and right of C).
MC = Fixing moment at C.
Fig. 25.8
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mid of span AB
=
2 2
1 2 (6)
8 8
wl ×
= = 9 kN-m
=
12 6
4 4
Wl ×
= = 18 kN-m
Now draw the μ-diagram with the help of above bending moments as shown in Fig. 25.8 (b).
From the geometry of bending moment diagrams, we find that for the spans 0A and AB,
0 0a x = 0
1 1a x =
1
9.0 6 3 108
2
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
Similarly, for the span AB and BC,
1 1a x =
2
9.0 6 3 108
3
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
2 2a x =
1
18.0 6 3 162
2
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
and for span BC and CD, 3 3a x =
1
18.0 6 3 162
2
⎛ ⎞× × × =⎜ ⎟
⎝ ⎠
0 0a x = 0
M0 l0 + 2 MA (0 + l1) + M0 l1 =
0 0 1 1
0 1
6 6a x a x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
0 + 2 MA (0 + 6) + MB × 6 =
6 108
0
6
×⎛ ⎞
+⎜ ⎟
⎝ ⎠
12 MA + 6 MB = – 108
or 2 MA + MB = – 18 ...(i)
Again using three moments equation for the spans AB and BC,
MA l1 + 2 MB (l1 + l2) + MC l2 =
1 1 2 2
1 2
6 6a x a x
l l
⎛ ⎞
+⎜ ⎟
⎝ ⎠
MA × 6 + 2 MB (6 + 6) + MC × 6 =
6 108 6 162
6 6
× ×⎛ ⎞
− +⎜ ⎟
⎝ ⎠
6 MA + 24 MB + 6 MC = – 270
or MA + 4 MB + MC = – 45 ...(ii)
Again using three moments equation for the spans BC and C 0,
MB l2 + 2 MC (l2 + l0) + MC l0 = –
0 02 2
2 0
66 a xa x
l l
⎛ ⎞
+⎜ ⎟
⎝ ⎠
MB × 6 + 2 MC (6 + 0) + 0 = –
6 162
0
6
×⎛ ⎞
+⎜ ⎟
⎝ ⎠
6 MB + 12 MC = – 162
or MB + 2 MC = – 27 ...(iii)
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Solving equations (i), (ii) and (iii)
MA = – 5.25 kN-m,
MB = – 7.5 kN-m
and MC = – 9.75 kN-m
Reactions at the supports
RC = Reaction at C.
Taking moments about B and equating the same,
RA × 6 + MB = MA + (2 × 6 × 3)
RA × 6 – 7.5 = – 5.25 + 36 = 30.75
∴ RA =
30.75 7.5
6
+
= 6.375 kN Ans.
Again taking moments about B and equating the same,
RC × 6 + MB = MC + 12 × 3
RC × 6 – 7.5 = – 9.75 + 36 = 26.25
∴ RC =
26.25 7.5
6
+
= 5.625 kN Ans.
and RB = (2 × 6 + 12) – (6.375 + 5.625) = 12 kN Ans.
Now draw the shear force diagram and elastic curve as shown in Fig. 25.8 (c) and 25.8 (d).
EXERCISE 25.1
1. A continuous beam is simply supported over three spans, such that AB = 8 m, BC = 12 m and
CD = 5 m. It carries uniformly distributed load of 4 kN/m in span AB, 3 kN/m in span BC and 6
kN/m in span CD. Find the moments over the supports B and C.
[Ans. – 35.9 kN-m ; – 31.0 kN-m]
2. A simply supported beam ABC is continuous over two spans AB and BC of 6 m and 5 m respec-
tively. The span AB is carrying a uniformly distributed load of 2 kN/m and the span BC is
carrying a point load of 5 kN at a distance of 2 m from B. Find the support moment and the
reactions. [Ans. – 7.1 kN-m ; 4.82 kN ; 11.6 kN ; 0.58 kN]
3. A continuous beam ABC is fixed at A and is simply supported at B and C. The span AB is 6 m
and carries a uniformly distributed load of 1 kN/m. The span AC is 4 m and carries a uniformly
distributed load of 3 kN/m. Determine the fixed end moments.
[Ans. MA = 2.143 kN-m ; MB = 4.714 kN-m ; MC = 0]
4. A continuous beam ABCD is simply supported over three spans of 5 m, 5 m and 4 m respec-
tively. The first two spans are carrying a uniformly distributed load of 4 kN/m, whereas the last
span is carrying a uniformly distributed load of 5 kN/m. Find the support moments at B and C.
[Ans. – 10.37 kN-m ; – 8.52 kN-m]
tively. The beam carries point loads of 90 kN and 80 kN at 2 m and 8 m from the support A and
a uniformly distributed load 30 kN/m over the span CD. Find the moments and reactions at the
supports. [Ans. 68.4 kN-m ; 44.8 kN-m ; 48.6 kN-m ; 90.1 kN ; 98.5 kN]
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25.7. Continuous Beams with End Span Overhanging
Sometimes, a continuous beam is overhanging, at its one or both ends. In such a case, the over-
hanging part of the beam behaves like a cantilever. The fixing moments on the end support may be
found out by the cantilever action of the overhanging part of the beam.
EXAMPLE 25.5. A beam ABCD 9 m long is simply supported at A, B and C, such that the
length AB is 3 m, length BC is 4.5 m and the overhung CD is 1.5 m. It carries a uniformly
distributed load of 1.5 kN/m in span AB and a point load of 1 kN at the free end D. The moments
of inertia of the beam in span AB and CD is I and that in the span BC is 2I. Draw the bending
moment and shear force diagrams for the beam.
SOLUTION. Given : Length AB (l1) = 3 m ; Length BC (l2) = 4.5 m ; Length CD (l3) = 1.5 m ;
Uniformly distributed load in AB (w) = 1.5 kN/m ; Point load at D = 1 kN ; Moment of inertia of
lengths AB and CD (IAB) = I and moment of inertia of length BC (IBC) = 2 I.
Bending moment diagram
Let *MA = Fixing moment at A,
MC = Fixing moment at C.
mid of span AB
=
2 2
1 1.5 (3)
8 8
wl ×
= = 1.69 kN-m
From the geometry of the figure, we find that the dixing moment at C,
MC = – 1.0 × 1.5 = – 1.5 kN-m
Fig. 25.9
Now draw the μ-diagram, with the help of above bending moments as shown in Fig. 25.9 (b).
From the geometry of the above bending moment diagram, we find that for the spans AB and BC,
1 1a x =
2
1.69 3 1.5 5.07
3
× × × =
* Since the beam is simply supported at A, therefore fixing moment MA will be zero.
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1 1 2 2
1 1 2 2
2A B C
l l l l
M M M
I I I I
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
=
1 1 2 2
1 1 2 2
6 6a x a x
I l I l
⎡ ⎤
− +⎢ ⎥
⎣ ⎦
3 4.5 4.5
0 2 1.5
2 2BM
I l l
⎛ ⎞
+ + + ×⎜ ⎟
⎝ ⎠
=
6 5.07
3l
×
−
×
10.5 6.75
2
BM
I I
− =
10.14
I
or 10.5 MB – 3.375 = – 10.14
∴ MB =
10.14 3.375
10.5
− +
= – 0.65 kN-m
Now complete the final bending moment diagram as shown in Fig. 25.9 (b).
Shear force diagram
RC = Reaction at C,
Taking moments at B,
RA × 3 – (1.5 × 3 × 1.5) = – 0.65 ...(ä MB = – 0.65 kN-m)
∴ RA =
0.65 6.75 6.1
3 3
− +
= = 2.03 kN
Again taking moments about B,
RC × 4.5 – (1 × 6) = – 0.65 ...(ä MB = – 0.65 kN-m)
∴ RC =
0.65 6 5.35
4.5 4.5
− +
= = 1.19 kN
and RB = [(3 × 1.5) + 1] – (2.03 + 1.19) = 2.28 kN
Now complete the shear force diagram as shown in Fig. 25.9 (c).
EXAMPLE 25.6. A beam ABCDE has a built-in support at A and roller supports at B, C and
D, DE being an overhung. AB = 7 m, BC = 5 m, CD = 4 m and DE = 1.5 m. The values of moment
of inertia of the section over each of these lengths are 3I, 2I, I and I respectively. The beam
carries a point load of 10 kN at a point 3 m from A, a uniformly distributed load of 4.5 kN/m over
whole of BC and a concentrated load of 9 kN in CD, 1.5 m from C and another point load of 3 kN
at E, the top of overhung as shown in Fig. 25.10.
Fig. 25.10
Determine (i) moments developed over each support and (ii) draw diagram for the entire beam,
stating values at salient points.
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SOLUTION. Given : Length AB (l1) = 7 m ; Length BC (l2) = 5 m ; Length CD (l3) = 4 m ; Length
DE (l4) = 1.5 m ; Moment of inertia of length AB (IAB) = 3 I ; Moment of inertia for length BC (IBC)
= 2 I ; Moment of inertia of length CD (ICD) = IDE = I ; Point load at F (W1) = 10 kN ; Uniformly
distributed load between B and C (w2) = 4.5 kN/m ; Point load at G (W2) = 9 kN and point load at E
= 3 kN.
(i) Moments developed over each support
Fig. 25.11
Since the beam is fixed at A, therefore, assume a zero span to the left of A.
Let *M0 = Fixing moment at left hand support of zero span,
MB = Fixing moment at B,
MC = Fixing moment at C and
MD = Fixing moment at D.
First of all, let us consider the beam AB as a simply supported beam. Therefore bending moment
under the 10 kN load
=
1 1 1
1
10 3 4
7
W a b
l
× ×
= = 17.14 kN-m
Similarly, bending moment under the 9 kN load in span CD
=
3 3 3
3
9 1.5 2.5
4
W a b
l
× ×
= = 8.44 kN-m
and bending moment at the mid of the span BC
=
2 2
2 2 4.5 (5)
8 8
w l ×
− = = 14.06 kN-m
From the geometry of the figure, we find that the fixing moment at D, due to load at E,
MD = – 3 × 1.5 = – 4.5 kN-m
Now draw μ-diagram with the help of above bending moments as shown in Fig. 25.11 (b). From
the geometry of the above bending moment diagram, we find that for the span 0A and AB,
0 0a x = 0
* Since there is a zero span on the left to A, therefore the fixing moment M0 will be zero.
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and 1 1a x =
2 41 1 3
4 17.14 3 17.14 4 220
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × + × × + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
Similarly, for the spans AB and BC,
1 1a x =
3 21 1 4
3 17.14 4 17.14 3 200
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × + × × + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
and 2 2a x =
2
3
× 5 × 14.06 × 2.5 = 117.2
Similarly, for the spans BC and CD,
2 2a x =
2
3
× 5 × 14.06 × 2.5 = 117.2
and 3 3a x =
2 2.51 1 1.5
2.5 8.44 1.5 8.44 2.5
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞× × × + × × +⎜ ⎟⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎝ ⎠⎣ ⎦
=36.6
0 0 1 1
0
0 0 1 1
2 A B
l l l l
M M M
I I I I
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
= – 0 0 1 1
0 0 1 1
6 6a x a x
I l I l
⎛ ⎞
+⎜ ⎟
⎝ ⎠
7 7
0 2
3 3A BM M
I I
+ × + =
6 220
0
3 7I
×⎛ ⎞
− +⎜ ⎟×⎝ ⎠
14 7
3 3
A BM M
I I
+ =
1320
3 7I
−
×
or 2 MA + MB = – 26.94 ...(i)
Again using three moments equation for the span AB and BC,
1 1 2 2
1 1 2 2
2A B C
l l l l
M M M
I I I I
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
= 1 1 2 2
1 1 2 2
6 6a x a x
I l I l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
7 7 5 5
2
3 3 2 2A B CM M M
I I I I
⎛ ⎞+ + +⎜ ⎟
⎝ ⎠
=
6 200 6 117.2
3 7 2 5I I
× ×⎛ ⎞
− +⎜ ⎟× ×⎝ ⎠
57 29
3 3 2
CA B MM M
I I I
+ + =
171.4 140.6
3 2I I
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
14 MA + 58 MB + 15 MC = – 764.6 ...(ii)
Again using three moments equation for the spans BC and CD,
3 32 2
2 2 3 3
2B C D
l ll l
M M M
I I I I
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
= 3 32 2
2 2 3 3
66 a xa x
I l I l
⎛ ⎞
+⎜ ⎟
⎝ ⎠
5 5 4 4
2 4.5
2 2B CM M
I I I I
⎛ ⎞+ + −⎜ ⎟
⎝ ⎠
=
6 117.2 6 36.6
2 5 4I I
× ×⎛ ⎞
− +⎜ ⎟× ×⎝ ⎠
265 18
2 2
CB MM
I I I
+ − =
140.6 54.9
2 I I
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
5 MB + 26 MC = – 214.4 ...(iii)
Now solving equations (i), (ii) and (iii), we get
MA = – 8.76 kN-m Ans.
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MB = – 9.41 kN-m Ans.
MC = – 6.44 kN-m Ans.
Now complete the bending moment diagram as shown in Fig. 25.11 (b)
25.8. Continuous Beams with a Sinking Support
Sometimes, one of the supports of a continuous beam sinks down due to loading with respect to
the other supports which remain at the same level. The sinking of a support effects the moments at the
supports.
Fig. 25.12
Now consider a continuous beam ABC fixed at A and C and supported at B as shown in Fig.
25.12. Let the support B* sink down through a more distance than the support C sinks down from its
original position (or in other words from the support A),
Let δA = Height of support A from B
and δC = Height of support C from B.
We have already discussed in Art. 25.3 that in the span AB
dEI1 [l1 iB – yB] = 1 1 1 1a x a x+ ′ ′
or EI1 [l1 iB – dA] = 1 1 1 1a x a x+ ′ ′
We have also discussed that
1 1a x′ + ′ = (MA + 2 MB)
2
1
6
l
∴ EI1 [l1 iB + δA) =
2
1
1 1 ( 2 )
6A B
l
a x M M+ +
EI1 l1 iB + EI1 δA =
2
1
1 1 ( 2 )
6A B
l
a x M M+ +
1
A
B
E
Ei
l
δ
+ = 1 1 1
1 1 1
( 2 )
6A B
a x l
M M
I l I
+ +
* It is also possible if all three supports sink down. But the support B sinks down more than A and C.
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or EiB = 1 1 1
1 1 1 1
( 2 )
6
A
A B
a x l E
M M
I l I l
δ
+ + −
Similarly, in the span BC taking C as the origin and x positive to the left,
Ei′E = 2 2 2
2 2 2 2
( 2 )
6
C
C B
Ea x l
M M
I l I l
δ
+ + −
Since iB is equal to – i′B, therefore EiB is equal to – Ei′B.
or 1 1 1
1 1 1 1
( 2 )
6
A
A B
a x l E
M M
I l I l
δ
+ + − = – 2 2 2
2 2 2 2
( 2 )
6
C
C B
Ea x l
M M
I l I l
δ⎡ ⎤
+ + −⎢ ⎥
⎣ ⎦
1 2 2 2
1 1 2 2
2 2
6 6 6 6A B C B
l l l l
M M M M
I I I I
+ + + = 1 1 2 2
1 1 2 2 2 2
CA Ea x a x E
I l I l l l
δδ⎛ ⎞
− + + +⎜ ⎟
⎝ ⎠
1 1 2 2
1 1 2 2
2A B C
l l l l
M M M
I I I I
⎛ ⎞
+ + +⎜ ⎟
⎝ ⎠
= 1 1 2 2
1 1 1 2 1 2
66 6 6 CA Ea x a x E
I l I l l l
δδ⎛ ⎞
− + + +⎜ ⎟
⎝ ⎠
NOTES: 1. If the moment of inertia of the beam is constant, then
MA l1 + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 2 1 2
66 6 6 CA EIa x a x EI
l l l l
δδ⎛ ⎞
− + + +⎜ ⎟
⎝ ⎠
2. The above formula has been derived by taking δA and δC as positive. But while solving the numericals on
sinking supports, care should always be taken to use the proper sign. The following guide rules should be
kept in mind for the purpose:
(i) The three moments equation always refer to two adjacent spans. The reference level should always be
taken as that of the common support.
(ii) The sign of δ for the left and right support, should then be used by comparing its height with the
central support (positive for higher and negative for lower), e.g., Consider a continuous beam ABCD
in which let the support B sink down by an amount equal to δ. The three supports, namely A, C and D
will remain at the same level. Now, while using three moments equation for the spans AB and BC, the
values of δA and δC will be positive (because both the supports A and C are at higher level than that of
B). But while using three moments equation for the spans BC and CD, the value of δB will be negative
(because the support B is at a lower level than that of C) and the value of δD will be zero (because the
support D is at the same level as that of C).
EXAMPLE 25.7. A continuous beam ABC, shown in Fig. 25.13 carries a uniformly distributed
load of 50 kN on AB and BC. The support B sinks by 5 mm below A and C and the values of EI is
constant throughout the beam.
Fig. 25.13
Taking E = 200 GPa and I = 332 × 10
6
mm
4
, find the bending moment at supports A and B and
draw the bending moment diagram.
SOLUTION. Given : Length AB (l1) = 4 m ; Length BC (l2) = 3 m ; Uniformly distributed load over
AC (w) = 50 kN-m ; Sinking of support of (δB) = – 5 mm = – 0.005 m or δA = δC = + 0.005 m ; Modulus
of elasticity (E) 200 GPa = 200 × 10
6
kN/m
2
and moment of inertia (I) = 332 × 10
6
mm
4
= 332 × 10
–6
m
4
.
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Fig. 25.14
Since the beam is fixed at A, therefore let us assume a zero span to the left of A,
MB = Fixing moment at B, and
*MC = Fixing moment at C,
First of all, consider the beam AB as a simply supported. Therefor bending moment at the mid of
the span AB
=
2 2
1 50 (4)
8 8
wl ×
= = 100 kN-m
=
2 2
2 50 (3)
8 8
wl ×
= = 56.25 kN-m
Now draw μ-diagram, with the help of above bending moments as shown in Fig. 25.14 (b). From
the geometry of the above bending moment diagram, we find that for the spans 0A and AB,
0 0a x = 0
and 1 1a x =
2 1600
100 4 2 533.3
3 3
× × × = =
1 1a x =
2 1600
100 4 2 533.3
3 3
× × × = =
and 2 2a x =
2
56.25 3 1.5 168.75
3
× × × =
M0 l0 + 2 MA (0 + l1) + MB l1 = – 0 0 01 1
0 1 1 2
6 66 6 Ba x EIa x EI
l l l l
δ δ⎡ ⎤
+ + +⎢ ⎥
⎣ ⎦
* Since the beam is simply supported at C, therefore fixing moment C will be zero.
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0 + 2 MA (0 + 4) + (MB × 4) =
6 533.3
0
4
×⎡ ⎤
− +⎢ ⎥⎣ ⎦
6 6
6 (200 10 ) (332 10 ) ( 0.005)
0
4
−
× × × × −
+ +
8 MA + 4 MB = – 800 – 498 = – 1298
∴ 2 MA + MB = – 324.5 ...(i)
MA l1 + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 2 1 2
6 6 6 6a x a x EI A EI C
l l l l
⎡ ⎤ δ δ
− + + +⎢ ⎥
⎣ ⎦
(MA × 4) + 2 MB (4 + 3) + 0 =
6 533.3 6 168.75
4 3
× ×⎡ ⎤
− +⎢ ⎥⎣ ⎦
6 6
6 (200 10 ) (332 10 ) ( 0.005)
4
−
× × × × +
+
6 6
6 200 10 332 10 ( 0.005)
3
−
× × × × +
+
4 MA + 14 MB = – (800 + 337.5) + 498 + 664 = 24.5
∴ 2 MA + 7 MB = 12.25 ...(ii)
Solving equations (i) and (ii),
MA = – 190.3 kN-m Ans.
and MB = 56.1 kN-m Ans.
NOTES: 1. While considering the spans 0A and AB, the assumed support 0 and fixed support A are at the same level.
But the support B has sunk down. Therefore the value of δ0 is taken as zero and that for δB as negative.
2. While considering the spans AB and BC, the supports A and C are at the same level. But the support
B has sunk down. Therefore the values δA and δC are taken as positive.
EXAMPLE 25.8. A continuous beam is built-in at A and is carried over rollers at B and C as
shown in Fig. 25.15. AB = BC = 12 m.
Fig. 25.15
It carries a uniformly distributed load of 3 kN/m over AB and a point load of 24 kN in BC, 4 m
from the support B, which sinks 30 mm. The values of E and I are 200 GPa and 0.2 × 109
respec-
tively and uniform throughout. Calculate the support moments and draw bending moment diagram
and shear force diagram, giving critical values. Also draw the deflected shape of the centre line of
the beam.
SOLUTION. Given : Length AB (l1) = 12 m ; Length BC (l2) = 12 m ; Uniformly distributed loas
in AB (w) = 3 kN/m ; Point loas at D (W) = 24 kN ; Sinking of support B (δB) = – 30 mm = – 0.03 m
or δA = δC = + 0.03 m ; modulus of elasticity (E) = 200 GPa = 200 × 10
6
kN-m
2
and sinking of
support B (I) = 0.2 × 10
9
mm
4
= 0.2 × 10
–3
m
4
.
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Fig. 25.16
Since the beam is fixed at A, therefore let us assume a zero span to the left of A,
Support moments at A, B and C
Let MA = Support moment at A,
MB = Support moment at B, and
MC = Support moment at C,
mid of span AB
=
2 2
1 3 (12)
8 8
wl ×
= = 54.0 kN-m
Similarly, bending moment under the 24 kN load
=
2
24 4 8
12
Wab
l
× ×
= = 64.0 kN-m
Now draw μ-diagram with the help of above bending moments as shown in Fig. 25.16 (b).
From the geometry of the above bending moment diagram, we find that for the spans 0A and AB,
0 0a x = 0
and 1 1a x =
2
3
× 54 × 12 × 6 = 2592
1 1a x =
2
3
× 54 × 12 × 6 = 2592
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and 2 2a x =
2 81 1 4
64 8 64 4 8 2560
2 3 2 3
⎡ × ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × + × × + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦
M0 l0 + 2 MA (l0 + l1) + MB l1 =
0 0 01 1
0 1 0 1
6 66 6 Ba x EIa x EI
l l l l
δ δ⎡ ⎤ ⎡ ⎤
− + + +⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
0 + 2 MA (0 + 12) + MB × 12 =
6 3
6 2592 6 (200 10 (0.2 10 ) ( 0.03)
0 0
12 12
−⎡ ⎤× × × × × × −⎡ ⎤
− + + +⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
24 MA + 12 MB = – 1296 – 600 = – 1896
∴ 2 MA + MB = – 158 ...(i)
MA l1 + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 1 1 2
66 6 6 CA EIa x a x EI
l l l l
δδ⎡ ⎤ ⎡ ⎤
− + + +⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
MA × 12 + 2 MB (12 + 12) + 0 =
6 2592 6 2560
12 12
× ×⎡ ⎤
− +⎢ ⎥⎣ ⎦
6 3
6 (200 10 ) (0.2 10 ) 0.03
12
−⎡⎛ ⎞× × × × ×
− ⎢⎜ ⎟⎜ ⎟
⎢⎝ ⎠⎣
6 3
6 (200 10 ) (0.2 10 ) 0.03
12
− ⎤⎛ ⎞× × × × ×
+ ⎥⎜ ⎟⎜ ⎟
⎥⎝ ⎠⎦
12 MA + 48 MB = – (1296 + 1280) + 600 + 600 = – 1376
∴ 3 MA + 12 MB = – 344 ...(ii)
Solving eqauations (i) and (ii)
MA = – 73.9 kN-m Ans.
MB = 10.2 kN-m Ans.
MC = 0 Ans.
Shear force diagram
RC = Reaction at C.
– 10.2 = RC × 12 – (24 × 4) ...(ä MB = – 10.2 kN-m)
∴ RC =
10.2 96
12
− +
= 7.15 kN
Now taking moments about A,
– 73.9 = 7.15 × 24 + RB × 12 – (24 × 16) – (3 × 12 × 6)
...(ä MA = – 73.9 kN-m)
= 12 RB – 428.4
∴ RB =
73.9 428.4
12
− +
= 29.54 kN
and RA = (3 × 12 + 24) – (7.15 + 29.54) = 23.31 kN
Now complete the Shear Force diagram as shown in Fig. 25.16 (c).
The elastic curve i.e., deflected shape of the centre line of the beam is shown in Fig. 25.16 (d).
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25.9. Continuous Beams Subjected to a Couple
Sometimes, a continuous beam is subjected to a couple in one (or more) of the spans. Such a
beam is also analyses in the similar manner. The couple will cause negative moments in one part and
positive moments in the other part of the span. While taking moment of the bending moment diagram,
due care should be taken for the positive and negative bending moments.
EXAMPLE 25.9. A continuous beam ABC of constant moment of inertia is simply supported
at A, B and C. The beam carries a central point load of 4 kN in span AB and a central clockwise
couple of moment 30 kN-m in span BC as shown in Fig. 25.17.
Fig. 25.17
Find the support moments and plot the shear force and bending moment diagrams.
SOLUTION. Given : Length AB (l1) = 10 m ; Length BC (l2) = 15 m ; Load at D (w) = 4 kN and
couple at E (μ) = 30 kN-m.
Support moments
Let *MA = Support moment at A,
MB = Support moment at B and
*MC = Support moment at C.
First of all, consider the beam AB and BC as a simply supported beam. Therefore bending mo-
ment at D,
MD =
1 4 10
4 4
Wl ×
= = 10 kN-m
We know that the **moment just on the right side of E
=
30
2 2
μ
= = 15.0 kN-m
and **moment just on the left side of E
=
30
2 2
μ
− = − = 15 kN-m
Now draw the -diagram with the help of above bending moments as shown in Fig. 25.18 (b).
From the geometry of the bending moment diagrams, we find that for spans AB and BC,
1 1a x =
1
2
× 10 × 10 × 5 = 250
2 2a x = 0 (ä + B.M. = – B.M.)
MA l1 + 2 MB (l1 + l2) + MC l2 = 1 1 2 2
1 2
6 6a x a x
l l
⎛ ⎞
− +⎜ ⎟
⎝ ⎠
* Since the beam is simply supported at A and C, therefore fixing moment MA and MC will be zero.
** For details, please refer to Art. 13.18.
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Fig. 25.18
0 + 2 MB (10 + 15) + 0 =
6 250
0
10
×⎛ ⎞
− +⎜ ⎟
⎝ ⎠
∴ 50 MB = – 150
or MB = – 3 kN-m
Shear force diagram
RC = Reaction at C.
RA × 10 – 4 × 5 = – 3 ...(ä MB = – 3 kN-m)
∴ RA =
3 20
10
− +
= 1.7 kN
Similarly, RC × 15 – 30 = – 3 ...(ä MB = – 3 kN-m)
∴ RC =
3 30
15
− +
= 1.8 kN
and RB = 4 – (1.7 + 1.8) = 0.5 kN
Now draw the shear force diagram as shown in Fig. 25.18 (c)
EXERCISE 25.2
1. A continuous beam ABCD is fixed at A and simply supported at B and C, the beam CD is
overhanging. The spans AB = 6 m, BC = 5 m and overhanging CD = 2.5 m. The moment of
inertia of the span BC is 2 l and that of AB and CD is l. The beam is carrying a uniformly
distributed load of 2 kN/m over the span AB, a point load of 5 kN in BC at a distance of 3 m
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from B and a point load of 8 kN at the free end.
Determine the fixing moments at A, B and C and draw the bending moment diagram.
[Ans. 8.11 kN-m ; 1.79 kN-m ; 20.0 kN-m]
2. A beam ABCD is continuous over three spans AB = 8 m, BD = 4 m and CD = 8 m. The beam AB
and BC is subjected to a uniformly distributed load of 1.5 kN/m, whereas there is a central point
load of 4 kN in CD. The moment of inertia of AB and CD is 2 l and that of BC is I. The ends A
and D are fixed. During loading, the support B sinks down by 10 mm. Find the fixed end
moments. Take E = 200 GPa and l = 1600 × 10
6
mm
4
.
[Ans. – 16.53 kN-m ; – 30.66 kN-m ; – 77.33 kN-m ; – 21.33 kN-m]
3. A continuous beam ABCD 20 m long is supported at B and C and fixed at A and D. The spans
AB, BC and CD are 6 m, 8 m and 6 m respectively. The span AB carries a uniformly distributed
load of 1 kN/m, the span BC carries a central point load of 10 kN and the span CD carries a
point load of 5 kN/m at a distance of 3 m from C. During loading, the support B sinks by 10 mm.
Find the fixed end moments and draw the bending moment diagram. Take E = 200 GPa and I =
3 × 10
9
mm
4
. The moment of inertia of the spans AB and CD is I and that of BC is 2I.
QUESTIONS
1. What a continuous beam is subjected to a couple in a span, it will cause negative moment in one
part and positive moment in other part of the span.
2. Explain the theorem of three moments.
3. Prove the Clapeyron’s theorem of three moments.
4. How will you apply the theorem of three moments to a fixed beam?
5. Explain the effect on a continuous beam, when one of the intermediate supports sinks down.
6. Describe the effect of a couple acting in one of the spans of a continuous beam.
OBJECTIVE TYPE QUESTIONS
1. Fixing moment over a simply supported end is
(a) zero (b) negative (c) positive (d) infinity
2. When a continuous beam is fixed at the left end, then an imaginary span is taken to the left of the
beam. The support moment at the imaginary support is
(a) negligible (b) considerable (c) zero (d) calculated
3. If a continuous beam is fixed at its both ends, then imaginary support is
(a) not taken (b) taken on left side only
(c) taken on both the ends (d) taken on right side only.
4. If one of the span of a continuous beam is subjected to a clockwise couple, then
(a) the span will be subjected to positive moment.
(b) the span will be subjected to negative moment.
(c) entire beam will be subjected to positive moment.
(d) one part of the span is subjected to positive moment and the other part to negative moment.
ANSWERS
1. (a) 2. (c) 3. (d) 4. (d)
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Moment
Distribution Method
Contents
1. Introduction.
2. Sign Conventions.
3. Carry Over Factor.
4. Carry Over Factor for a Beam
Fixed at One End and Simply
Supported at the Other.
5. Carry Over Factor for a Beam,
Simply Supported at Both Ends.
6. Stiffness Factor.
7. Distribution Factors.
8. Application of Moment Distribution
Method to Various Types of
Continuous Beams.
9. Beams with Fixed End Supports.
10. Beams with Simply Supported
Ends.
11. Beams with End Span
Overhanging.
12. Beams With a Sinking Support.
26.1. Introduction
The moment distribution method, which was
first introduced by Prof. Hardy Cross in 1930, is
widely used for the analysis of all types of
indeterminate structures. In this method, all the
members of a structure are first assumed to be
fixed in position and direction and fixed end
moments due to external loads are obtained. Now
all the hinged joints are released, by applying an
equal and opposite moment and their effects are
evaluated on the opposite joints. The unbalanced
moment at a joint, is distributed in the two spans
in the ratio of their distribution factors. This
process is continued, till we reach the required
degree of precision.
26C h a p t e r
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Chapter 26 : Moment Distribution Method 625
26.2. Sign Conventions
Though different types of sign conventions are adopted by different authors in their books, yet
the following sign conventions, which are widely used and internationally recognised will be used in
this book.
1. All the clockwise moments at the ends are taken as positive.
2. All the anticlockwise moments at the ends are taken as negative.
26.3. Carry Over Factor
We have already discussed in Art. 26.1 that the moments are applied on all the end joints of a
structure, whose effects are evaluated on the other joints. The ratio of moment produced at a joint to
the moment applied at the other joint, without displacing it, is called carry over factor. Now, we shall
find out the value of carry over factor in the following two cases of beam:
1. When the beam is fixed at one end and simply supported at the other.
2. When the beam is simply supported at both the ends.
26.4. Carry Over Factor for a Beam Fixed at One End and
Simply Supported at the Other
Consider a beam AB fixed at A and simply supported at B. Let a clockwise moment be applied at
the support B of the beam as shown in Fig. 26.1.
Let l = Span of the beam,
μ = Clockwise moment applied at B (i.e., MB) and
MA = Fixing moment at A.
Since the beam is not subjected to any external loading, therefore the two reactions (R) must be
equal and opposite as shown in Fig. 26.1.
Fig. 26.1
Taking moments about A and equating the same,
R · l = MA + μ ...(i)
Now consider any section X, at a distance x from A. We know that the moment at X,
MX = MA – R · x
or
2
2
d y
EI
dx
= MA – R · x ...
2
2
d y
M EI
dx
⎛ ⎞
=⎜ ⎟⎜ ⎟
⎝ ⎠
∵
Integrating the above equation,
dy
EI
dx
=
2
1·
2A
Rx
M x C− +
where C1 is the first constant of integration. We know that when x = 0, then
dy
dx
= 0. Therefore C1 = 0.
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or
dy
EI
dx
=
2
·
2
Rx
MA x − ...(ii)
Integrating the above equation once again,
EI · y =
2 3
2
·
2 6
AM x Rx
C− +
where C2 is the second constant of integration. We know that when x = 0, then y = 0. Therefore C2 = 0.
or EI · y =
2 3
·
2 6
AM x Rx
− ...(iii)
We also know that when x = l, then y = 0. Therefore substituting these values in equation (iii),
0 =
2 3
· ·
2 6
AM l R l
−
∴
3
6
Rl
=
2
·
2
AM l
or R · l = 3MA
Substituting the value of in equation (i),
3MA = MA + μ
or MA =
2 2
BMμ
= ...(iv)
∴ A
B
M
M
=
1
2
It is thus obvious, that carry over factor is one-half in this case,
We see from equation (ii) that
dy
EI
dx
= MA · x
2
2
Rl
−
Now for slope at B, substituting x = l in the above equation,
EI · iB = MA · l
2
2
Rl
− ...(ä R · l = 3MA)
= MA · l
3
·
2 AM l−
=
·
2 4
AM l lμ
− = − ... 2AM
μ⎛ ⎞=⎜ ⎟
⎝ ⎠
∵
∴ iB =
4
l
EI
μ
−
=
4
l
EI
μ
or μ =
4 · BEI i
l
26.5. Carry Over Factor for a Beam Simply Supported at Both Ends
Consider a beam AB simply supported at A and B. Let a clockwise moment be applied at the
support B of the beam as shown in Fig. 26.2.
...(Minus sign means that the tangent at
B makes an angle with AB in the
negative or anticlockwise direction)
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Fig. 26.2
Let l = Span of the beam, and
μ = Clockwise moment at B.
Since the beam is simply supported at A, therefore there will be no fixing moment at A. Moreover,
as the beam is not subjected to any external loading, therefore the two reactions must be equal and
opposite as shown in Fig. 26.2.
Taking moments about A,
R · l = μ ...(i)
Now consider any section X, at a distance x from A. We know that the moment at X,
MX = – R · x
or
2
2
d y
EI
dx
= – R · x ...
2
2
d y
M EI
dx
⎛ ⎞
=⎜ ⎟⎜ ⎟
⎝ ⎠
∵
Integration the above equation,
dy
EI
dx
=
2
1
2
Rx
C− + ...(ii)
where C1 is the first constant of integration. Integrating the above equation once again,
EI · y =
3
1 2
6
Rx
C x C− + +
where C2 is the second constant of integration. We know that when x = 0, then y = 0. Therefore C2 = 0.
or EI · y =
3
1
6
Rx
C x− + ...(iii)
We also know that when x = l, then y = 0. Therefore substituting these values in the above
equation,
0 =
3
1
6
Rl
C− + l
∴ C1 =
2
6 6
lRl μ
= ...(ä R · l = μ)
Substituting this value of C1 in equation (ii),
dy
EI
dx
=
2 2
2 6 2 6
l lRx Rlx
l
μ μ
− + = − +
=
2
2 6
x l
l
μ μ
− + ...(ä R · l = μ)
Now for slope at B, substituting x = l in the above equation,
EI · iB =
2
2 6 2 6 3
l l l l l
l
μ μ μ μ μ
− + = − + = −
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∴ iB =
3
l
EI
μ
−
=
3
l
EI
μ
∴ μ =
3 · BEI i
l
26.6. Stiffness Factor
It is the moment required to rotate the end, while acting on it, through a unit angle without
translation of the far end. We have seen in Art. 26.4 that the moment on a beam having one end fixed
and the other freely supported,
μ =
4 · BEI i
l
∴ Stiffness factor for such a beam (substituting iB = 1),
k1 =
4EI
l
Similarly, we have seen in Art. 26.5 that the moment on a beam having simply supported ends,
μ =
3 · BEI i
l
∴ Stiffness factor for such a beam (substituting iB = 1),
k2 =
3EI
l
26.7. Distribution Factors
Sometimes, a moment is applied on a structural joint to produce rotation, without the translation
of its members. This moment is distributed among all the connecting members of the joint in the
proportion of their stiffness.
Consider four members OA, OB, OC and OD meeting at A. Let the members OA and OC be fixed
at A and C, whereas the members OB and OD be hinged at B and D. Let the joint O be subjected to a
moment μ as shown in Fig. 26.3.
Fig. 26.3
Let l1 = Length of the member OA,
I1 = Moment of inertia of the member OA,
... (Minus sign means that the tangent
at B makes an angle with AB in the
negative or anticlockwise direction)
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E1 = Modulus of elasticity of the member OA,
l1, I2, E2 = Corresponding values for the member OB,
l3, I3, E3 = Corresponding values for the member OC,
and l4, I4, E4 = Corresponding values for the member OD.
A little consideration will show that as a result of the moment μ, each member gets rotated
through some equal angle. Let this angle through which each member is rotated be θ.
We know that the stiffness of member OA,
k1 = 1 1
1
4E I
l
...(ä End A is fixed)
Similarly, k2 = 2 2
2
3E I
l
...(ä End B is hinged)
and k3 = 3 3
3
4E I
l
...(ä End C is fixed)
and k4 = 4 4
4
3E I
l
...(ä End D is hinged)
Now total stiffness of all the members,
k = k1 + k2 + k3 + k4
and total moment applied at the joint,
μ = kθ
∴ Moment on the member OA,
μ1 = k1θ
Similarly, μ2 = k2θ : μ3 = k3θ and μ4 = k4θ
∴ 1μ
μ
= 1 1k k
k k
θ
=
θ
Similarly, 2μ
μ
= 2k
k
; 3μ
μ
= 3k
k
and 4 4k
k
μ
=
μ
∴ μ1 = k1 k
μ
Similarly, μ2 = 2k
k
μ
; μ3 = 3k
k
μ
and μ4 = 4k
k
μ
The quantities 31 2
, ,
kk k
k k k
and 4k
k
are known as distribution factors for the members OA, OB, OC
and OD respectively. The moments μ1, μ2, μ3 and μ4 are known as distributed moments.
EXAMPLE 26.1. Five members OA, OB, OC, OD and OE meeting at O, are hinged at A and
C and fixed at B, D and E. The lengths of OA, OB, OC, OD and OE are 3 m, 4 m, 2 m, 3 m and 5
m and their moments of inertia are 400 mm4
, 300 mm4
, 200 mm4
, 300 mm4
and 250 mm4
respectively. Determine the distribution factors for the members and the distributed moments,
when a moment of 4000 kN-m is applied at O.
SOLUTION: Given: Length OA = 3 m ; Length OB = 4 m ; Length OC = 2 m ; Length OD = 3 m;
Length OE = 5 m ; Moment of inertia of OA = 400 mm
4
; Moment of inertia of OB = 300 mm
4
;
Moment of inertia of OC = 200 mm
4
; Moment of inertia of OD = 300 mm
4
; Moment of inertia of OE
= 250 mm
4
and moment on D = 4000 kN-m.
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We know that stiffness factor for OA,
kA =
3 4003
4
EEI
I
× ×
= = 400 E ...(ä Member is hinged at A)
Similarly, kB =
4 3004
4
EEI
I
× ×
= ...(ä Member is fixed at B)
Now complete the column for stiffness for all the members, keeping in mind whether the member
is hinged or fixed at the end. Now find out the distribution factor and distribution moments for each
member as shown in the above chart.
Now from the above chart, we find that the distribution factors for OA, OB, OC, OD and OE are
1 3 3 1
, , ,
4 16 16 4
and 1
8
respectively Ans.
The moment of 4000 kN-m applied at the joint O will be distributed among the members as
obtained from the above in the following table:
Member Length M.I. Stiffness Distribution Distributed
(m) (mm
4
) (k) factor moments N-m
OA 3 400
3 400
3
E ×
= 400
400 1
1600 4
E
E
= 1
4000
4
× = 1000
OB 4 300
4 300
4
E ×
= 300
300 3
1600 16
E
E
=
3
4000
16
× = 750
OC 2 200
3 200
2
E ×
= 300
300 3
1600 16
E
E
=
3
4000
16
× = 750
OD 3 300
4 300
3
E ×
= 400
400 1
1600 4
E
E
= 1
4000
4
× = 1000
OE 5 250
4 250
5
E ×
= 200
200 1
1600 8
E
E
=
1
4000
8
× = 500
Total ΣΣΣΣΣ = 1600E ΣΣΣΣΣ = 400
Thus distributed moments for OA, OB, OC, OD and OE are 1000, 750, 750, 1000 and 500 N-m
respectively. Ans.
26.8. Application of Moment Distribution Method to Various Types
of Continuous Beams
In the previous articles, we have studied the principles of the moment distribution method. First
of all, all the supports are assumed to be clamped and fixing moments are found out for all the spans.
The unbalanced moment at a support is distributed among the two spans in the ratio of their stiffness
factors and their effects are evaluated on the opposite joints. This process is continued till we reach
the desired degree of precision. Though the moment distribution method has very wide applications
yet in the proceeding articles, we shall discuss its application to the following types of beams only:
1. Beams with fixed end supports,
2. Beams with simply supported ends,
3. Beams with end span overhanging and
4. Beams with a sinking support.
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26.9. Beams with Fixed End Supports
Sometimes, a continuous beam is fixed at its one or both ends. In such a case, the fixed end
moments are obtained and the unbalanced moment is distributed in two spans in the ratio of their
distribution factors.
EXAMPLE 26.2. Evaluate the bending moment and shear force diagrams of the beam shown
in Fig. 26.4.
What are the reactions at the supports?
Fig. 26.4
* SOLUTION. Given: Length AB (l1) = 6 m ; Length BC (l2) = 6 m ; Uniformly distributed load in
AB (w) = 2 kN/m and point load at D (W) = 12 kN.
First of all, let us consider the continuous beam ABC to be split into two fixed beams AB and BC
as shown in Fig. 26.5 (b) and (c).
We know that in span AB the fixing moment at A
=
2
12
wl
− ...(Minus sign due to anticlockwise)
Fig. 26.5
=
2
2 (6)
12
×
− = – 6.0 kN-m
* We have already solved this question by Theorem of three moments in Example 25.4.
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and fixing moment at B =
2
12
wl
+ ...(Plus sign due to clockwise)
=
2
2 (6)
12
×
= 6.0 kN-m
Similarly, in span BC the fixing moment at B
=
8
Wl
=
12 6
8
×
− = – 9.0 kN-m
and fixing moments at C =
8
Wl
=
12 6
8
×
= 9.0 kN-m
Now let us find out the distribution factors at B. From the geometry of the figure, we find that the
stiffness factor for BA,
kB =
4 4 2
6 3
EI EI EI
l
= = ...(ä The beam is fixed at A)
Similarly, kB =
4 4 2
6 3
EI EI EI
l
= = ...(ä The beam is fixed at C)
∴ Distribution factors for the members BA and BC
=
2
3
2 2
3 3
EI
EI EI
+
and
2
3
2 2
3 3
EI
EI EI
+
=
1
2
and
1
2
Fig. 26.6
Now draw the beam and fill up the distribution factors and initial moments as shown in Fig. 26.6.
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We find that the fixing moment at B in span AB, is + 6.0 kN-m and that in span BC, is – 9.0 kN-
m. Thus there is an unbalanced moment equal to (– 9.0 + 6.0) = – 3.0 kN-m. Now, distribute this
unbalanced moment (equal to – 3.0 kN-m) into the spans AB and BC in the ratio of their distribution
factors i.e., + 1.5 kN and + 1.5 kN. Now *carry over the effects of these distributed moments at A and
D equal to
1
1.5
2
× = + 0.75 kN. Then distribute the unbalanced moment at B (In this case, there is no
carry over moment from A or D at B. So the distribution of moment at B is zero). Now, find out the
final moments at A, B and C in the spans AB and BC, by algebraically adding the respective values as
shown in the last column of the table.
We know that the bending moment at the centre of the span AB, treating it as a simply supported
beam
=
22
2 (6)
8 8
wl ×
= = 9.0 kN-m
Similarly, bending moment under the load in the span BC
=
12 6
4 4
Wl ×
= = 18.0 kN-m
Now complete the final bending moment **diagram as shown in Fig. 26.7 (b).
RC = Reaction at C.
Fig. 26.7
* As per At. 26.4 the carry over factor is
1
2
.
** Though the moment at B in the span AB, is positive and moment at C in the span BC is positive in the
table, yet these moments are taken as negative in the bending moment diagram. The reason for the same
is that these points the moments tend to bend the beam with convexity upwards.
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Taking moments at B and equating the same,
RA × 6 + MB = MA + 2 × 6 × 3
RA × 6 – 7.5 = – 5.25 + 36 = 30.75
∴ RA =
30.75 7.5
6
+
= 6.375 kN Ans.
Again taking moments about B and equating the same,
RC × 6 + MB = MC + 12 × 3
RC × 6 – 7.5 = – 9.75 + 36 = 26.25
∴ RC =
26.25 7.5
6
+
= 5.625 kN Ans.
and RB = (2 × 6 + 12) – (6.375 + 5.625) = 12.0 kN Ans.
Now complete the shear force diagram and elastic curve as shown in Fig. 26.7 (c) and (d).
26.10. Beams with Simply Supported Ends
Sometimes, a continuous beam is simply supported curve on one or both of its ends. We know
that the fixing moment on a simply supported end is zero.Therefore in such a case, the simply supported
ends are released by applying equal and opposite moments and their effects are carried over on the
opposite joints. It may also be noted that no moment is carried over from the opposite joint to the
simply supported end.
EXAMPLE 26.3 A continuous beam ABC 10 m long rests on three supports A, B and C at the
same level and is loaded as shown in Fig. 26.8.
Fig. 26.8
Determine the moments over the beam and draw the bending moment diagram. Also calculate
the reactions at the supports and draw the shear force diagram.
*SOLUTION. Given: Length AB (l1) = 6 m ; Length BC (l2) = 4 m ; Load at D (W) = 3 kN ;
Distance AD (a) = 2 m ; Distance DB (b) = 4 m and uniformly distributed load BC (w) = 1 kN/m.
First of all, assume the continuous beam ABC to split up into two fixed beams AB and BC as
shown in Fig. 26.9 (b) and (c).
In span AB, the fixing moment at A
=
2
2
Wab
l
=
2
2
3 2 (4) 8
3(6)
× ×
− = − = – 2.67 kN-m
and fixing moment at B
=
2
2
Wa b
l
=
2
2
3 (2) 4 4
3(6)
× ×
+ = + = + 1.33 kN-m
* We have already solved this question by Theorem of three moments in Example 25.1.
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Fig. 26.9
Similarly, in span BC, the fixing moment at B
=
2
12
wl
=
2
1 (4) 4
12 3
×
− = − = – 1.33 kN-m
and fixing moment at C
=
2
12
wl
=
2
1 (4) 4
12 3
×
+ = + = + 1.33 kN-m
Now let us find out the distribution factors at B. From the geometry of the figure, we find that the
kBA =
3 3
6 2
EI EI EI
l
= = ...(ä The beam is hinged at A)
Similarly, stiffness factor for BC,
kBC =
3 3 3
4 4
EI EI EI
l
= = ...(ä The beam is hinged at C)
Distribution factors for BA and BC
= 2
3
2 4
EI
EI EI
+
and
3
4
3
2 4
EI
EI EI
+
=
2
5
and
3
5
Now draw the beam and fill up the distribution factors and fixing moments as shown in Fig.
26.10 (a). Now release the ends A and C (because of simply supported ends) by applying equal and
opposite moments. Now *carry over the moments from A to B and C to B. After adding the carry over
* As per article 26.5, the carry over factor is
1
2
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moments at B, we find that the moment at B in span AB is + 2.66 kN-m and in span BC is – 2.00. Now
distribute the unbalanced moment of 0.66 kN-m (i.e., 2.66 – 2.00 = 0.66 kN-m) into the two spans of
AC in the ratio of their distribution factors, i.e., – 0.26 kN-m and – 0.40 kN. Now find out the final
moments by adding all the values.
We know that the bending moment under 3 tonnes load in the span AB treating it as a simply
supported
=
3 2 4
6
Wab
l
× ×
= = 4.0 kN-m
Similarly, bending moment at the mid of the span BC due to uniformly distributed load
=
22
1 (4)
8 8
wl ×
= = 2.0 kN-m
Now prepare the following table.
Fig. 26.10
Now complete the final bending moment diagram as shown in Fig. 26.11(b).
Shear force diagram
RC = Reaction at C.
RA × 6 – 3 × 4 = – 2.4 ...(ä MB = 2.4 kN-m)
∴ RA =
2.4 12.0 9.6
6 6
− +
= = 1.6 kN Ans.
Similarly,
RC × 4 – 4 × 2 = – 2.4 ...(ä MB = – 2.4 kN-m)
∴ RC =
2.4 8.0 5.6
4 4
− +
= = 1.4 kN Ans.
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and RB = (3 + 4) – (1.6 + 1.4) = 4.0 kN Ans.
Now draw shear force diagram as shown in Fig. 26.11 (c).
Fig. 26.11
EXERCISE 26.1
1. A continuous beam ABC is fixed at A and is simply supported at B and C. The span AB is 6 m
and carries a uniformly distributed load of 1 kN/m. The span AC is 4 m and carries a uniformly
distributed load of 3 kN/m. Determine the fixed end moments.
[Ans. ( = 2.143 kN-m ; MB = 4.714 kN-m ; MC = 0)]
2. A simply supported beam ABC is continuous over two spans AB and BC of 6 m and 5 m respec-
tively. The span AB is carrying a uniformly distributed load of 2 kN-m and the span BC is
carrying a point load of 5 kN at a distance of 2 m from B. Find the support moment at B.
[Ans. 7.1 kN-m]
3. A continuous beam ABCD is simply supported over three spans, such that AB = 8 m BC = 12 m
and CD = 5 m. It carries uniformly distributed load of 4 kN/m in span AB, 3 kN/m in span BC
and 6 kN/m in span CD. Find the moments over the supports B and C.
[Ans. 35.9 kN-m ; 31.0 kN-m]
tively. The beam carries point loads of 90 kN and 80 kN at 2 m and 8 m from the support A and
a uniformly distributed load of 30 kN-m over the span CD. Find the moments and reactions at
the supports. [Ans. 68.4 kN-m ; 44.8 kN ; 48.6 kN ; 94.1 kN ; 98.5 kN]
26.11. Beams with End Span Overhanging
Sometimes, a beam is overhanging at its one or both the end supports. In such a case, the bending
moment at the support near the overhanging end will be due to the load over the overhanging portion
and will remain constant, irrespective of the moments on the other supports. It is thus obvious that the
distribution factors over the support having one span overhanging will be 1 and 0. Moreover, this
support is considered as a simply supported for the purpose of calculating distribution factors in the
span, adjoining the overhanging span.
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EXAMPLE 26.4. A beam ABCD 9 m long is simply supported at A, B and C such that the span
AB is 3 m, span BC is 4.5 m and the overhanging CD is 1.5 m. It carries a uniformly distributed load
of 1.5 kN/m in span AB and a point load of 1 kN at the free end B. The moment of inertia of the beam
in span AB is I and that in the span BC is 2I. Draw the B.M. and S.F. diagrams of the beam.
*SOLUTION. Given : Length AB (l1) = 3 m ; Length BC (l2) = 4.5 m ; Length CD (l3) = 1.5 m ;
Moment of inertia for AB (IAB) = ICD = I ; Moment of inertia for BC = (IBC) = 2I ; Uniformly
distributed load between A and B (w) = 1.5 kN/m and point load at D (W) = 1 kN.
First of all, let us assume the continuous beam ABCD to be split up into fixed beams AB, BC and
cantilever CD.
In span AB the fixing moment at AP
=
22
1.5 (3)
12 12
wl ×
− = − = – 1.125 kN-m
=
22
1.5 (3)
12 12
wl ×
+ = + = + 1.125 kN-m
Since the span BC is not carrying any load, therefore fixing moment at B and C will be zero. The
moment at C for the cantilever CD will be 1 × 1.5 = –1.5 kN-m.
Now let us find out the distribution factors at B and C. From the geometry of the figure, we find
that the stiffness factor for BA,
kBA =
3 3
1
3
EI EI
EI
l
= =
Similarly, kBC =
3 23 6 4
4.5 4.5 3
E lEI EI EI
l
×
= = =
Distribution factor for BA and BC
=
2
4
1
3
EI
EI
EI +
and
4
3
4
1
3
EI
EI
EI +
=
3
7
and
4
7
We know that distribution factors for CA and CD will be 1 and 0, because the beam is overhanging
at C.
Now prepare the following table.
A B C D
3
7
4
7
1 0
– 1.125 + 1.125 0 0 – 1.5 Fixed end moments
+ 1.125 0 0 + 1.5 0 Release A and balance C
+ 0.562 + 0.75 0 0 Carry over
0 + 1.687 + 0.75 + 1.5 – 1.5 Initial moments
– 1.037 – 1.40 – – Distribute
0 + 0.65 – 0.65 + 1.5 – 1.5 Final moments
* We have already solved, this question by Theorem of three momens in Example 25.5
...(ä The end A is simply
supported at A)
...(ä Beam is over
handing beyond C)
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We know that the bending moment in the mid of span AB due to uniformly distributed load, by
considering it as a simply supported beam
=
22
1.5 (3)
8 8
wl ×
= = 1.69 kN-m
Now complete the final bending moment diagram as shown in Fig. 26.12(b).
Shear force diagram
RC = Reaction at C.
Taking moments at B
RA × 3 – (1.5 × 3 × 1.5) = – 0.65 ...(ä MB = – 0.65 kN-m)
Fig. 26.12
or RA =
0.65 6.75
3
− +
= 2.03 kN
Again taking moments about B,
RC × 4.5 – (1 × 6) = – 0.65 ... (ä MB = – 0.65 kN-m)
or RC =
0.65 6
4.5
− +
= 1.19 kN
∴ RB = (3 × 1.5 + 1) – (2.03 + 1.19) = 2.28 kN
EXAMPLE 26.5. A continuous beam ABCDE, with uniform flexural rigidity throughout has
roller supports at B, C and D, a built-in support E and an overhang AB as shown in Fig. 26.13.
Fig. 26.13
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It carries a uniformly distributed load of intensity of 2 kN/m on AB and another of intensity of
3 kN/m over BCDE. In addition to it, a point load of 8 kN is placed midway between C and D. The
span lengths are AB = 2 m, BC = CD = DE = 5 m. Obtain the support moments by the moment
distribution method and sketch the B.M. diagram giving values at salient points.
SOLUTION. Given: Length AB (l1) = 2 m ; Length BC (l2) = Length CD (l3) = Length DE (l4) = 5
m ; Uniformly distributed load on AB (w1) = 2 kN/m ; Uniformly distributed load in BC, CD, DE (w2)
= 3 kN/m and point load at F (W) = 8 kN.
Support moments
First of all, let us assume the continuous beam ABCDE to be split up into cantilever AB and fixed
beams BC, CD and DE.
We know that the bending moment at B for the cantilever AB,
MB =
1
1 1
2
2 2
2 2
l
w l × = × × = 4 kN-m
In span BC, fixing moment at B
=
2 2
2 3 (5)
12 12
w l ×
− = − = – 6.25 kN-m
and fixing moment at C =
2 2
2 3 (5)
12 12
w l ×
+ = + = + 6.25 kN-m
In span CD, the fixing moment at C
=
2 2
2
212
w l Wab
l
⎡ ⎤
− +⎢ ⎥
⎢ ⎥⎣ ⎦
=
2 2
2
3 (5) 8 2.5 (2.5)
12 5
⎡ ⎤× × ×
− +⎢ ⎥
⎢ ⎥⎣ ⎦
kN-m
= – 11.25 kN-m
and fixing moment at D =
2 2
2
212
w l Wa b
l
⎡ ⎤
+ +⎢ ⎥
⎢ ⎥⎣ ⎦
=
2 3
2
3 (5) 8 2.5 (2.5)
12 5
⎡ ⎤× × ×
+ +⎢ ⎥
⎢ ⎥⎣ ⎦
kN-m
= + 11.25 kN-m
In span DE, the fixing moment at D
=
2 2
2 3 (5)
12 12
w l ×
− = − = – 6.25 kN-m
and fixing moment at E
=
2 2
2 3 (5)
12 12
w l ×
+ = = + 6.25 kN-m
Now let us find out the distribution factors at B, C and D. From the geometry of the figure, we
find that the distribution factors for BA and BC will be 0 and 1.
Stiffness factor for CB,
kCB =
3EI
l
=
3
5
EI
...(∵ The beam is overhanging beyond C)
and kCD =
4EI
l
=
4
5
EI
...(∵ The beam is continuous at D)
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Distribution factors for CB and CD
=
3
5
3 4
5 5
EI
EI EI
+
and
4
5
3 4
5 5
EI
EI EI
+
= 3
7
and 4
7
Similarly, stiffness factor for DC,
kCD =
4EI
l
=
4
5
EI
...(∵ The beam is continuous at C)
and kDE =
4EI
l
=
4
5
EI
...(ä The beam is fixed at E)
Distribution factors for DC and DE
=
4
5
4 4
5 5
EI
EI EI
+
and
4
5
4 4
5 5
EI
EI EI
+
= 1
2
and 1
2
Now prepare the following table:
A B C D E
0 1
3
7
4
7
1
2
1
2
+ 4.00 – 6.25 + 6.25 – 11.25 + 11.25 – 6.25 + 6.25 Fixed end moments
+ 2.25 Balance B
+ 1.13 Cary over
+ 4.00 – 4.00 + 7.38 – 11.25 + 11.25 – 6.25 + 6.25 Initial moments
+ 1.66 + 2.21 – 2.50 – 2.50 Distribute
– 1.25 + 1.10 – 1.25 Carry over
+ 0.54 + 0.71 – 0.55 – 0.55 Distribute
– 0.28 + 0.36 – 0.28 Cary over
+ 0.12 + 0.16 – 0.18 – 0.18 Distribute
– 0.09 + 0.08 – 0.09 Carry over
+ 0.04 + 0.05 – 0.04 – 0.04 Distribute
+ 4.00 – 4.00 + 9.74 – 9.74 + 9.52 – 9.52 + 4.63 Final moments
We know that the bending moment in the middle of span BC, by considering it as a simply
supported beam
=
2
2
8
w l
=
2
3 (5)
8
×
= 9.38 kN-m
Similarly, bending in the middle of span CD, by considering it as a simply supported beam
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=
2
2
8 1
w l Wab
+ =
2
3 (5) 8 2.5 2.5
8 5
× × ×
+ = 19.38 kN-m
and bending moment in the middle of span DE, by considering it as a simply supported beam
=
2
2
8
w l
=
2
3 (5)
8
×
= 9.38 kN-m
Fig. 26.14
EXAMPLE 26.6. A beam ABCDE has a built-in support A and roller supports at B, C and D,
DE being an overhung. AB = 7 m, BC = 3 m, CD = 4m and DE = 1.5 m. The values of I, the
moment of inertia of the section, over each of these lengths are 3I, 2I, I and I respectively. The
beam carries a point load of 10 kN at a point 3 m from A, a uniformly distributed load of 4.5 kN/
m over whole of BC and a concentrated load of 9 kN in CD 1.5 m from C and another point load
of 3 kN at E the top of overhang as shown in Fig. 26.15.
Fig. 26.15
Determine (i) moments developed over each of the supports A, B, C and D and (ii) draw
bending moment diagram for the entire beam, stating values at salient points.
*SOLUTION. Given : Length AB (l1) = 7 m ; Length BC (l2) = 5 m ; Length CD (l3) = 4 m ; Length
DE (l4) = 1.5 m ; Moment of inertia for AB (IAB) = 3I ; Moment of inertia for BC (IBC) = 2I ; Moment
of inertia for CD (ICD) = IDE = I ; Point load at F (W1) = 10 kN ; Uniformly distributed load on BC
(w1) = 4.5 kN/m ; Point load at G (W2) = 9 kN and point load at E = 3 kN.
Moments developed over each of the support
First of all, let us assume the continuous beam ABCDE to be split up into fixed beams AB, BC,
CD and cantilever DE.
* This question is also solved in Example 25.6.
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In span AB, fixing moment at A
=
2 2
1
2 2
10 3 (4)
(7 )
W ab
l
× ×
− = − = – 9.80 kN-m
2 2
1
2 2
10 (3) 4
(7)
W a b
l
× ×
+ = = 7.35 kN-m
In span BC, fixing moment at A
=
22
4.5 (5)
12 12
wl ×
− = − = – 9.38 kN-m
22
4.5 (5)
12 12
wl ×
+ = − = 9.38 kN-m
In span CD, fixing moment at C
=
2 2
2
2 2
9 1.5 (2.5)
(4)
W ab
l
× ×
− = − = – 5.27 kN-m
and fixing moment at D =
2 2
2
2 2
9 (1.5) 2.5
(4)
W a b
l
× ×
+ = − = 3.16 kN-m
and the moment at D, for the cantilever at DE
= 3 × 1.5 = – 4.5 kN-m
Now let us find out the distribution factors at B, C and D. From the geometry of the figure, we
find that the stiffness factor for BA,
kBA =
4 34 12
7 7
E IEI EI
l
×
and kBC =
4 24 8
5 5
E IEI EI
l
×
= = ...(ä The beam is continuous at C)
=
12
7
12 8
7 5
EI
EI EI
+
and
8
5
12 8
7 5
EI
EI EI
+
=
15
29
and 14
29
Similarly, stiffness factor for CB,
kCB =
4 24 8
5 5
E IEI EI
l
×
= = ...(ä The beam is continuous at B)
and kCD =
3 3
4
EI EI
l
= ...(ä The beam is overhanging beyond D)
Distribution factors for CB and CD
=
8
5
8 3
5 4
EI
EI EI
+
and
3
4
8 3
5 4
EI
EI EI
+
=
32
47 and 15
47
Distribution factors for DC and DE will be 1 and 0, because the beam is overhanging at D.
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A B C D E
15
29
14
29
32
47
15
47
1 0
– 9.80 + 7.35 – 9.38 + 9.38 – 5.27 + 3.16 – 4.5 Fixed end moments
+ 1.34 Balance D
+ 0.67 Carry over
– 9.80 + 7.35 – 9.38 + 9.38 – 4.60 + 4.5 – 4.5 Initial moments
+ 1.05 + 0.98 – 3.25 – 1.53 Distribute
+ 0.53 – 1.63 + 0.49 0 Carry over
+ 0.84 + 0.79 – 0.33 – 0.16 Distribute
+ 0.42 – 0.17 + 0.40 0 Carry over
+ 0.09 + 0.08 – 0.27 – 0.13 Distribute
+ 0.05 – 0.13 + 0.04 0 Carry over
0.07 + 0.06 – 0.03 – 0.01 Distribute
+ 0.04 – 0.02 + 0.03 0 Carry over
0.01 0.01 – 0.02 – 0.01 Distribute
– 8.76 9.41 – 9.41 + 6.44 – 6.44 + 4.5 – 4.5 Final moments
The moments developed at each of the support are given in the above table. Ans.
We know that the bending moment under the 10 kN load in span AB, treating it as a simply
supported
Fig. 26.16
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= 1 10 3 4
7
W ab
l
× ×
= = 17.14 kN-m
=
22
4.5 (5)
8 8
wl ×
= = 14.06 kN-m
and bending moment under the 9 kN load in span CD
= 2 9 1.5 2.5
4
W ab
l
× ×
= = 8.44 kN-m
26.12. Beams with a Sinking Support
Sometimes, one of the supports of a continuous beam sinks down with respect to others, as a
result of loading. As a result of sinking, some moments are caused on the two supports, in addition to
the moments due to loading.
Fig. 26.17
Consider a beam ABC simply supported at A, B and C subjected to any loading. As a result of this
loading, let the support B sink down by an amount equal to δ as shown in Fig. 26.17 (b). Now assume
the beam ABC to be split up into two beams AB and BC as shown in Fig. 26.17 (c) and (d).
Now from the geometry of the beam AB, we find that the moment caused at A and B due to
sinking of the support B
= 2
6EI
l
δ
− ...(Minus sign due to right support sinking down)
Similarly, moment caused at B and C of the beam BC due to sinking of the support B
= 2
6EI
l
δ
+ ...(Plus sign due to left support sinking down)
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EXAMPLE 26.7. A continuous beam ABC shown in Fig. 26.18 carries a uniformly distributed
load of 50 kN/m on AB and BC. The support B sinks by 5 mm below A and C and the value of EI
is constant throughout the beam.
Fig. 26.18
Taking E = 200 GPa and I = 332 × 10
6
mm
4
, find the bending moment at supports A and B and
draw the bending moment diagram.
*Solution. Given : Length AB (l1) = 4 m ; Length BC (l2) = 3 m ; Uniformly distributed load (w)
= 50 kN/m ; Sinking of support B (δB) = –5 mm = – 0.005 m or δA = δC = + 0.005 m ; Modulus of
elasticity (E) = 200 GPa = 200 × 10
6
kN/m
2
and moment of inertia (I) = 332 × 10
6
mm
4
= 332 × 10
–6
m
4
.
Bending moment at supports
First of all, let us assume the continuous beam ABC to be split up into fixed beams AB and BC. In
span AB, fixing moment at A due to loading
=
22
50 (4)
12 12
wl ×
− = − = – 66.7 k-m
=
22
50 (4)
12 12
wl ×
+ = = 66.7 kN-m
Now in span AB moment at A due to sinking of support B
= 2
6EI
l
δ
− ...(Minus sign due to right support sinking)
=
6 6
2
6 (200 10 ) (332 10 ) 0.005
(4)
−
× × × × ×
− kN-m
= – 124.5 kN-m
Similarly, moment at B, due to sinking of support B
= – 124.5 kN-m
∴ Total moment at A in span AB
= – 66.7 – 124.5 = – 191.2 kN-m
and total moment at B in span AB
= + 66.7 – 124.5 = – 57.8 kN-m
In span BC, fixing moment at B due to loading
=
22
50 (3)
12 12
wl ×
− = − = – 37.5 kN-m
22
50 (3)
12 12
wl ×
+ = = + 37.5 kN-m
Now in span BC, moment at B due to sinking of support B
= 2
6EI
l
δ
+ ...(Plus sign due to left support sinking)
* We have already solved this question by Theorem of there moments in Example 25.7.
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=
6 6
2
6 (200 10 ) (332 10 ) 0.005
(3)
−
× × × × ×
+ kN-m
= 221.3 kN-m
Similarly, moment at C, due to sinking of support B
= + 221.3 kN-m
∴ Total moment at B, in span BC
= – 37.5 – 221.3 = 183.8 kN-m
and total moment at C, in span BC
= + 37.5 – 221.3 = 258.8 kN-m
Now, let us find out the distribution factors at B. From the geometry of the figure, we find that the
kBA =
4 4
4
EI EI
l
= = 1 EI ...(ä The beam is fixed at A)
and kBC =
3 3
3
EI EI
l
= ...(ä The beam is simply suppoted at C)
Distribution factors for BA and BC will be
1
2
and
1
2
.
A B C
1
2
1
2
– 191.2 – 57.8 + 183.8 + 258.8 Fixed end moment
– 258.8 Release C
– 129.4 Carry over
– 191.2 – 57.8 + 54.4 0 Initial moments
+ 1.7 + 1.7 Distribute
+ 0.9 Carry over
0 0 0 0 Distribute
– 190.3 – 56.1 + 56.1 0 Final moments
The bending moments at the supports are given in the above table. Ans.
We know that the bending moment at the mid of the span AB, treating it as a simply supported
beam
=
22
50 (4)
8 8
wl ×
= = 100 kN-m
=
22
50 (3)
8 8
wl ×
= = 56.25 kN-m
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Fig. 26.19
EXAMPLE 26.8. A continuous beam is built-in at A and is carried over rollers at B and C as
shown in Fig. 26.20. AB = BC = 12 m.
It carries a uniformly distributed load of 3 kN/m over AB and a point load of 24 kN over BC,
4 m from the support B, which sinks 30 mm. Values of E and I are 200 GPa and 0.2 × 109
m4
respectively uniform throughout.
Calculate the support moments and draw bending moment diagram and shear force diagram,
giving critical values. Also draw the deflected shape of the centre line of the beam.
Fig. 26.20
*SOLUTION. Given : Length AB (l1) = 12 m ; Length BC (l2) = 12 m ; Uniformly distributed load
on AB (w) = 3 kN/m ; Point load at D (W) = 24 kN ; Sinking of support B (δB) = – 30 mm = – 0.03
m or δA = (δC) = + 0.03 m ; Modulus of elasticity (E) = 200 GPa = 200 × 106
kN-m2
and moment of
inertia (I) = 0.2 × 109
mm4
= 0.2 × 10–3
m4
.
Support moments at A, B and C
First of all, let us assume the continuous beam ABC to be split up into fixed beams AB and BC.
In span AB, the fixing moment at A due to loading
=
22
3 (12)
12 12
wl ×
− = − = – 36.0 kN-m
22
3 (12)
12 12
wl ×
+ = + = + 36.0 kN-m
Now in span AB, the moment at A due to sinking of support
* We have already solved this question by Three moments theorem in Example 25.8.
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= 2
6EI
l
δ
− ... (Minus sign due to right support sinking)
=
6 3
2
6 (200 10 ) (0.2 10 ) 0.03
(12)
−
× × × × ×
− = – 50.0 kN-m
Similarly, moment at B due to sinking of support B
= – 50.0 kN-m
∴ Total moment at A in span AB
= – 36.0 – 5.00 = – 86.0 kN-m
and total moment at B in span AB
M = + 36.0 – 50.0 = – 14.0 kN-m
Now in span BC, the fixing moment at B due to loading
=
22
2 2
24 4 (8)
(12)
Wab
l
× ×
− = − = – 42.67 kN-m
22
2 2
24 (4) 8
(12)
Wa b
l
× ×
+ = = + 21.33 kN-m
Now in span BC, the moment at B, due to sinking of support B
= 2
6EI
l
δ
+ ... (Plus sign due to left support sinking)
=
6 3
2
6 (200 10 ) (0.2 10 ) 0.03
(12)
−
× × × × ×
+ = + 50.0 kN-m
Similarly, moment at C due to sinking of support B = + 50.0 kN-m
∴ Total moment at B in span BC
= – 42.67 + 50.0 = + 7.33 kN-m
and total moment at C in span BC
= + 21.33 + 50.0 = + 71.33 kN-m
Now, let us find out the distribution factors at B. From the geometry of the figure, we find that the
kBA =
4 4
12 3
EI EI EI
l
and kBC =
3 3
12 4
EI EI EI
l
= = ...(ä The beam is simply supported at C)
= 3
3 4
EI
EI EI
+
and 4
3 4
EI
EI EI
+
= 4
7
and 3
7
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A B C
4
7
3
7
– 86.00 – 14.00 + 7.33 + 71.33 Fixed end moment
– 71.33 Release C
– 35.67 Carry over
– 86.00 – 14.00 – 28.34 0 Initial moments
+ 24.20 + 18.14 Distribute
+ 12.10 Carry over
0 0 0 0 Distribute
– 73.9 + 10.2 – 10.2 0 Final moments
The bending moments at the supports are given in the above table. Ans.
From the geometry of the figure, we find that the bending moment at the mid of the span AB
treating it as a simply supported beam
=
22
3 (12)
8 8
wl ×
= = 54.0 kN-m
Fig. 26.21
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Similarly, bending moment under the 24 kN load
=
24 4 8
12
Wab
l
× ×
= = 64.0 kN-m
Now complete the bending moment as shown in Fig. 26.21 (b).
Shear force diagram
RC = Reaction at C.
Taking moments about B, and equating the same
RC × 12 – (24 × 4) = – 10.2 ...(ä MB = – 10.2 kN-m)
∴ RC =
10.2 96.0
12
− +
= 7.15 kN
Now taking moments about A, and equating the same
– 73.9 = (7.15 × 24) + RB × 12 – (24 × 16) – (3 × 12 × 6)
...(ä Bending moment A = – 73.9 kN-m)
= 12RB – 428.4
∴ RB =
73.9 428.4
12
− +
= 29.54 kN
and RA = (3 × 12 + 24) – (7.15 + 29.54) = 23.31 kN
The elastic curve i.e., deflected shape of the centre line of the beam is shown in Fig. 26.21 (d).
EXERCISE 26.2
1. A continuous beam ABCD is fixed at A and simply supported at B and C, the beam CD is
overhanging. The spans AB = 6 m, BC = 5 m and overhanging CD = 2.5 m. The moment of
inertia of the span BC is 2l and that of AB and CD is l. The beam is carrying a uniformly
distributed load 2 kN/m over the span AB, a point load of 5 kN in BC at a distance of 3 m from
B, and a point load of 8 kN at the free end.
Determine the fixing moments at A, B and C and draw the bending moment diagram.
[Ans. 8.11 kN-m ; 1.79 kN-m ; 20.0 kN-m]
2. A beam ABCD is continuous over three spans AB = 8 m, BC = 4 m and CD = 8 m. The beam AB
and BC is subjected to a uniformly distributed load of 1.5 kN/m, whereas there is a central point
load of 4 kN in CD. The moment of inertia of AB and CD is 2I and that of BC is I. The end A and
D are fixed. During loading the support A sinks down by 10 mm. Find the fixed end moments.
Take E = 200 Gpa and I = 16 × 10
6
mm
4
.
[Ans. –16.53 kN-m ; – 30.66 kN-m ; – 77.33 kN-m ; – 21.33 kN-m]
3. A continuous beam ABCD 20 m long is supported at B and C and fixed at A and D. The spans
AB, BC and CD are 6 m, 8 m and 6 m respectively. The span AB carries a uniformly distributed
load of 1 kN/m, the span BC carries a central point load of 10 kN and the span CD carries a
point load of 5 kN at a distance of 3 m from C. During loading, the support B sinks by 10 mm.
Find the fixed end moments and draw the bending moment diagram. Take E = 10 mm and I = 3
× 10
9
mm
4
. The moment of inertia of the spans AB and CD is I and that of BC = 2I.
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QUESTIONS
1. Define the term ‘carry over factor’. Derive a relation for the stiffness factor for a beam simply
supported at its both ends.
2. What do you understand by the term ‘distribution factor’? Discuss its importance in the method
of moment distribution.
3. Explain the procedure for finding out the fixed end moments in:
(a) beams with fixed end supports. (b) beams with simply supported ends.
(c) beams with end span overhanging. (d) beams with a sinking support.
OBJECTIVE TYPE QUESTIONS
1. Stiffness factor for a beam fixed at one end and freely supported at the other is
(a)
3EI
l
(b)
4EI
l
(c)
6EI
l
(d)
8EI
l
2. Stiffness factor for beam simply supported at both the ends is“
(a)
3EI
l
(b)
4EI
l
(c)
6EI
l
(d)
8EI
l
3. If the end B of a continuous beam ABC sinks down, then the moment at A will be
(a) zero (b) negative (c) positive (d) infinity
ANSWERS
1. (b) 2. (a) 3. (b)
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Torsion of
Circular Shafts
Contents
1. Introduction.
2. Assumptions for Shear Stress in a
Circular Shaft Subjected to Torsion.
3. Torsional Stresses and Strains.
4. Strength of a Solid Shaft.
5. Strength of hollow shaft.
6. Power Transmitted by a Shaft.
7. Polar Moment of Inertia.
8. Replacing a Shaft.
9. Shaft of Varying Section.
10. Composite Shaft.
11. Strain Energy due to Torsion.
13. Combined Bending and Torsion
14. Combined Bending and torsion
along with Axial Thurst
15. Shaft Couplings.
16. Design of Bolts.
17. Design of Keys.
27.1. Introduction
In workshops and factories, a turning force
is always applied to transmit energy by rotation.
This turning force is applied either to the rim of a
pulley, keyed to the shaft or at any other suitable
point at some distance from the axis of the shaft.
The product of this turning force and the distance
between the point of application of the force and
the axis of the shaft is known as torque, turning
moment or twisting moment.And the shaft is said
to be subjected to torsion. Due to this torque,
every cross-section of the shaft is subjected to
some shear stress.
27.2. Assumptions for Shear
Stress in a Circular Shaft
Subjected to Torsion
Following assumptions are made, while
27C h a p t e r
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finding out shear stress in a circular shaft subjected to torsion:
1. The material of the shaft is uniform throughout.
2. The twist along the shaft is uniform.
3. Normal cross-sections of the shaft, which were plane and circular before the twist, remain plane
and circular even after the twist.
4. All diameters of the normal cross-section, which were straight before the twist, remain straight
with their magnitude unchanged, after the twist.
A little consideration will show that the above assumptions are justified, if the torque applied is
small and the angle of twist is also small.
27.3. Torsional Stresses and Strains
Fig. 27.1
Consider a circular shaft fixed at one end and subjected to a torque at the other end as shown in
Fig. 27.1.
Let T = Torque in N-mm,
l = Length of the shaft in mm and
R = Radius of the circular shaft in mm.
As a result of this torque, every cross-section of the shaft will be subjected to shear stresses. Let
the line CA on the surface of the shaft be deformed to CA′ and OA to OA′ as shown in Fig. 27.1.
Let ∠ACA′ = φ in degrees
∠AOA′ = θ in radians
τ = Shear stress induced at the surface and
C = Modulus of rigidity, also known as torsional rigidity of the
shaft material.
We know that shear strain = Deformation per unit length
=
AA
l
′
= tan θ
= φ ...(φ being very small, tan φ = φ)
We also know that the arc AA′ = R · θ
∴ φ =
RAA
l l
′ ⋅ θ
= ...(i)
If τ is the intensity of shear stress on the outermost layer and C the modulus of rigidity of the
shaft, then
φ =
C
τ
...(ii)
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Chapter 27 : Torsion of Circular Shafts 655
From equations (i) and (ii), we find that
C
τ
=
R
l
⋅ θ
or
C
R l
τ ⋅θ
=
If τX be the intensity of shear stress, on any layer at a distance x from the centre of the shaft, then
X
x
τ
=
C
R l
⋅ θτ
= ...(iii)
27.4. Strength of a Solid Shaft
The term, strength of a shaft means the maximum torque or power, it can transmit. As a matter of
fact, we are always interested in calculating the torque, a shaft can withstand or transmit.
Let R = Radius of the shaft in mm and
τ = Shear stress developed in the outermost layer of the shaft in
N/mm
2
Consider a shaft subjected to a torque T as shown in Fig. 27.2. Now let us consider an element of
area da of thickness dx at a distance x from the centre of the shaft as shown in Fig. 27.2.
∴ da = 2πx · dx ...(i)
and shear stress at this section,
∴ τX =
x
R
τ × ...(ii)
where τ = Maximum shear stress.
∴ Turning force = Shear Stress × Area
= τx · dx
=
x
da
R
τ × ×
= 2 ·τ × π
x
x dx
R
=
22
·
x
x dx
R
π
We know that turning moment of this element,
dT = Turning force × Distance of element from axis of the shaft
=
2 32 2
x dx x x dx
R R
πτ πτ
⋅ = ⋅ ...(iii)
The total torque, which the shaft can withstand, may be found out by integrating the above
equation between 0 and R i.e.,
T =
3 2
0 0
2 2
R R
x dx x dx
R R
πτ πτ
⋅ = ⋅∫ ∫
=
4
3 3
0
2
4 2 16
R
x
R D
R
⎡ ⎤πτ π π
= τ ⋅ = × τ ×⎢ ⎥
⎣ ⎦
N-mm
where D is the diameter of the shaft and is equal to 2R.
EXAMPLE 27.1. A circular shaft of 50 mm diameter is required to transmit torque from one
shaft to another. Find the safe torque, which the shaft can transmit, if the shear stress is not to
exceed 40 MPa.
SOLUTION. Given: Diameter of shaft (D) = 50 mm and maximum shear stress (τ) = 40 MPa
= 40 N/mm
2
.
Fig. 27.2
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We know that the safe torque, which the shaft can transmit,
T =
3 3
40 (50)
16 16
D
π π
× τ × = × × N-mm
= 0.982 × 106
N-mm = 0.982 kN-m Ans.
EXAMPLE 27.2. A solid steel shaft is to transmit a torque of 10 kN-m. If the shearing stress
is not to exceed 45 MPa, find the minimum diameter of the shaft.
SOLUTION. Given: Torque (T) = 10 kN-m = 10 × 10
6
N-mm and maximum shearing stress (τ) = 45
MPa = 45 N/mm
2
.
Let D = Minimum diameter of the shaft in mm.
We know that torque transmitted by the shaft (T),
10 × 10
6
=
3 3
45
16 16
D D
π π
× τ × = × × = 8.836 D
3
∴ D
3
=
6
10 10
8.836
×
= 1.132 × 10
6
or D = 1.04 × 102
= 104 mm Ans.
27.5. Strength of a Hollow Shaft
It means the maximum torque or power a hollow shaft can transmit from one pulley to another.
Now consider a hollow circular shaft subjected to some torque.
Let R = Outer radius of the shaft in mm,
r = Inner radius of the shaft in mm, and
τ = Maximum shear stress developed in the outer most layer of
the shaft material.
Now consider an elementary ring of thickness dx at a
distance x from the centre as shown in Fig. 27.3.
We know that area of this ring,
da = 2πx · dx ...(i)
and shear stress at this section,
τX =
x
R
τ ×
∴ Turning force = Stress × Area
= τX · dx
... X
x
R
⎛ ⎞τ = τ ×⎜ ⎟
⎝ ⎠
∵
= 2
x
xdx
R
τ × × π ...(ä da = 2π xdx)
=
22
x dx
R
πτ
⋅ ...(ii)
We know that turning moment of this element,
dT = Turning force × Distance of element from axis of the shaft
=
2 32 2
x dx x x dx
R R
πτ πτ
⋅ ⋅ = ⋅ ...(iii)
Fig. 27.3
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The total torque, which the shaft can transmit, may be found out by integrating the above equation
between r and R.
∴ T =
3 32 2
R R
r r
x dx x dx
R R
πτ πτ
⋅ = ⋅∫ ∫
=
4 4 4 44
2 2
N-mm
4 4 16
R
r
R r D dx
R R D
⎛ ⎞ ⎛ ⎞⎡ ⎤ − −πτ πτ π
= = × τ ×⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎜ ⎟ ⎜ ⎟
⎣ ⎦ ⎝ ⎠ ⎝ ⎠
where D is the external diameter of the shaft and is equal to 2R and 2d is the internal diameter of the
shaft and is equal to 2r.
NOTE: We have already discussed in Art. 27.3 that the shear stress developed at a point is proportional to its
distance from the centre of the shaft. It is thus obvious that in the central portion of a shaft, the shear
stress induced is very small. In order to utilize the material to the fuller extent, hollow shafts are used.
EXAMPLE 27.3. A hollow shaft of external and internal diameter of 80 mm and 50 mm is
required to transmit torque from one end to the other. What is the safe torque it can transmit, if the
allowable shear stress is 45 MPa ?
SOLUTION. Given: External diameter (D) = 80 mm; Internal diameter (d) = 50 mm and allowable
shear stress (τ) = 45 MPa = 45 N/mm2
.
We know that torque transmitted by the shaft,
T =
4 4 4 4
(80) (50)
45
16 16 80
D d
D
⎡ ⎤ ⎡ ⎤− −π π
× τ × = × ×⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦
N-mm
= 3.83 × 106
N-mm = 3.83 kN-m Ans.
27.6. Power Transmitted by a Shaft
We have already discussed that the main purpose of a shaft is to transmit power from one shaft to
another in factories and workshops. Now consider a rotating shaft, which transmits power from one
of its ends to another.
Let N = No. of revolutions per minute and
T = Average torque in kN-m.
Work done per minute = Force × Distance = T × 2πN = 2πNT
Work done per second =
2
kN-m
60
NTπ
Power transmitted = Work done in kN-m per second
=
2
kW
60
NTπ
NOTE: If the torque is in the N-m, then work done will also be in N-m and power will be in watt (W).
EXAMPLE 27.4. A circular shaft of 60 mm diameter is running at 150 r.p.m. If the shear
stress is not to exceed 50 MPa, find the power which can be transmitted by the shaft.
Solution. Given: Diameter of the shaft (D) = 60 mm ; Speed of the shaft (N) = 150 r.p.m. and
maximum shear stress (τ) = 50 MPa = 50 N/mm
2
.
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T =
3 3
50 (60) N-mm
16 16
D
π π
× τ × = × ×
= 2.12 × 106
N-mm = 2.12 kN-m
and power which can be transmitted by the shaft,
P =
2 150 2.122
60 60
NT π × ×π
= = 33.3 kW Ans.
EXAMPLE 27.5. A hollow shaft of external and internal diameters as 100 mm and 40 mm is
transmitting power at 120 r.p.m. Find the power the shaft can transmit, if the shearing stress is
not to exceed 50 MPa.
SOLUTION. Given: External diameter (D) = 100 mm; Internal diameter (d) = 40 mm ; Speed of
the shaft (N) = 120 r.p.m. and allowable shear stress (τ) = 50 MPa = 50 N/mm
2
.
We know that torque the shaft can transmit,
T =
4 4 4 4
(100) (40)50
16 16 100
D d
D
⎡ ⎤− ⎡ ⎤π π −× τ × = × ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
N-mm
= 9.56 × 106
N-mm = 9.56 kN-m
and power the shaft can transmit,
P =
2 120 9.562
60 60
NT π × ×π
= = 120 kW Ans.
EXAMPLE 27.6. A solid circular shaft of 100 mm diameter is transmitting 120 kW at
150 r.p.m. Find the intensity of shear stress in the shaft.
SOLUTION. Given : Diameter of the shaft (D) = 100 mm ; Power transmitted (P) = 120 kW and
speed of the shaft (N) = 150 r.p.m.
Let T = Torque transmitted by the shaft, and
τ = Intensity of shear stress in the shaft.
We know that power transmitted by the shaft (P),
120 =
2 1502
60 60
TNT π × ×π
= = 15.7 T
∴ T =
120
15.7
= 7.64 kN-m = 7.64 × 10
6
N-mm
We also know that torque transmitted by the shaft (T),
7.64 × 10
6
=
3 3
(100)
16 16
D
π π
× τ × = × τ × = 0.196 × 10
6
τ
τ =
7.64
0.196
= 39 N/mm2
= 39 MPa Ans.
EXAMPLE 27.7. A hollow shaft is to transmit 200 kW at 80 r.p.m. If the shear stress is not to
exceed 60 MPa and internal diameter is 0.6 of the external diameter, find the diameters of the
shaft.
SOLUTION. Given : Power (P) = 200 kW ; Speed of shaft (N) = 80 r.p.m. ; Maximum shear stress
(τ) = 60 MPa = 60 N/mm2
and internal diameter of the shaft (d) = 0.6D (where D is the external
diameter in mm).
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T =
4 4 4 4
(0.6 )
60
16 16
D d D D
D D
⎡ ⎤ ⎡ ⎤− −π π
× τ × = × ×⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
N-mm
= 10.3D3
N-mm=10.3 ×10–6
D3
kN-m ...(i)
We also know that power transmitted by the shaft (P),
200 =
6 3
2 80 (10.3 10 )2
60 60
DNT
−
π × × ×π
= = 86.3 × 10
–6
D
3
∴ D
3
= 6
200
(86.3 10 )
−
×
= 2.32 ×10
6
mm
3
or D = 1.32 × 102
= 132 mm Ans.
and d = 0.6 D = 0.6 × 132 = 79.2 mm Ans.
EXAMPLE 27.8. A solid steel shaft has to transmit 100 kW at 160 r.p.m. Taking allowable
shear stress as 70 MPa, find the suitable diameter of the shaft. The maximum torque transmitted
in each revolution exceeds the mean by 20%.
SOLUTION. Given: Power (P) = 100 kW ; Speed of the shaft (N) = 160 r.p.m. ; Allowable shear
stress (τ) = 70 MPa = 70 N/mm2
and maximum torque (Tmax) = 1.2 T (where T is the mean torque).
Let D = Diameter of the shaft in mm.
We know that power transmitted by shaft (P),
100 =
2 1602
60 60
TNT π × ×π
= = 16.8 T
∴ T =
100
16.8
= 5.95 kN-m = 5.95 × 10
6
N-mm
and maximum torque, Tmax = 1.2T = 1.2 × (5.95 × 10
6
) = 7.14 × 10
6
N-mm
We also know that maximum torque (Tmax),
7.14 × 10
6
=
3 3
70
16 16
D D
π π
× τ × = × × = 13.7 D
3
∴ D3
=
6
7.14 10
13.7
×
= 0.521 × 106
or D = 0.8 × 10
2
= 80 mm Ans.
EXERCISE 27.1
1. A circular shaft of 80 mm diameter is required to transmit torque in a factory. Find the torque,
which the shaft can transmit, if the allowable shear stress is 50 MPa. (Ans. 5.03 kN-mm)
2. A solid steel shaft is required to transmit a torque of 6.5 kN-m. What should be the minimum
diameter of the shaft, if the maximum shear stress is 40 MPa? (Ans. 94 mm)
3. A solid shaft of 40 mm diameter is subjected to a torque of 0.8 kN-m. Find the maximum shear
stress induced in the shaft. (Ans. 63.7 MPa)
4. A hollow shaft of external and internal diameters of 60 mm and 40 mm is transmitting torque.
Find the torque it can transmit, if the shear stress is not to exceed 40 MPa. (Ans. 1.36 kN-m)
5. A circular shaft of 80 mm diameter is required to transmit power at 120 r.p.m. If the shear stress
is not to exceed 40 MPa, find the power transmitted by the shaft. (Ans. 50.5 kW)
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