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Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 1
1. A steel girder of I section supported over a span of
3.6 m has Ixx = 1,200 × 104 mm4. Determine the slope at
ends and deflection at the centre of the beam, if central
load is 16 kN, E = 200 GPa = 200 × 109 N/m2 = 200 ×
106 kN/m2.
Solution:
E = 200 × 106 kN/m2
I = 1,200 × 104 mm4
EI = 1,200 × 200 × 10-2 = 2,400 kNm2
W = 16 kN; L = 3.6m
2. A beam AB, 10 m long, carries point loads of 6 and 3
kN at C and D as shown in Fig. Determine support
reactions, deflection at C and D, and slope at
ends A and B, if EI is the flexural rigidity of the beam.
Solution:
Taking moments about A,
6 × 4 + 7 × 3 = 10RB
Reaction, RB = (24 + 21)/10 = 4.5 kN
Total load on beam = 6 + 3 = 9 kN
Reaction at A, RA = 9 − RB = 9 − 4.5 = 4.5 kN.
There are three portions, that is, AC, CD and DB in the
beam and if A is the origin then DB is the last portion.
Consider a section at a distance x from A in the
portion DB,
Bending moment, Mx = 4.5x − 6(x − 4) − 3(x − 7)
Integrating
Since the beam is not symmetrically loaded about its
centre, so we do not know where slope is zero.
By integrating eq (ii) we obtain,
EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 + C1x + C2
Boundary conditions:
x = 0 at end A, deflection y = 0
Moreover in portion AC, only the first term is valid and
the other two terms of equation are not valid.
So, 0 = 0.75 × 0, neglected term – neglected term + C1 ×
0 + C2
or, constant C2 = 0
So, EIy = 0.75x3 – (x – 4)3 – 0.5(x – 7)3 + C1x
At end B, x = 10 m, deflection y = 0, by substituting the
value
0 = 0.75 × 103 – 63 – 0.5 × 33 + 10C1
0 = 750 – 216 –13.5 + 10C1
Constant, C1 = –52.05
Finally, the equations are
EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 − 52.05x
Slope at end A, x = 0
EIiA = 2.25 × 0 – neglected terms –52.05
at end B, x = 10, so all the terms are valid
Deflection
At point C, x = 4 m, hence, the third term in the
equation is invalid.
EIyC = 0.75 × 43 − (4 − 4)3 − neglected term − 52.05 × 4
EIyC = 48 − 0 − 208.2 = −160.2
At point D, x = 7 m, and all the terms in the equation for
deflection are valid.
EIyD = 0.75 × 73 − (7 − 4)3 − 0.5(7 − 7)3 − 52.05 × 7 =
257.25 − 27 − 0 − 364.35 = −134 1
3. A beam ABCD, 6 m long hinged at end A and roller
supported at end D, is subjected to CCW moment of 10
kN m at point B and a point load of 10 kN at point C as
shown in Fig. Determine the deflection under load of
10 kN and slope at point B, by taking EI as flexural
rigidity of the beam.
Solution:
Taking moments at A,
10 + RD × 6 = 10 × 4
RD = 5kN ↑
Total load on the beam = 10 kN
Reaction at A, RA = 10 – 5 = 5kN ↑
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 2
There are three portions: AB, BC and CD in beam,
with A as the origin portion and CD is the last. Take a
section YY in portion CD at a distance x from A as
shown in the figure.
The equation of bending moment is
Note that 10 kN m is a moment and we cannot take
moment of moment, but 10 kN m is applied at B at a
distance of (x – 2) m from section YY. Moreover (x – 2)°
= 1, so only to locate the position of moment, the term
(x – 2)° is taken, that is, (x – 2) raised to power zero, this
term locates the position of moment applied at B.
Integrating we get
where C1 is a constant of integration, we do not know
where the slope is zero as the beam is not
symmetrically loaded.
Also by integrating we obtain
where C2 is another constant of integration.
y = 0 at end A, x = 0, only first two terms in between are
valid in portion AB, so
or, constant, C2 = 0.
y = 0 at end D, where x = 6 m, all the terms in the
equation are valid, so
Finally, the equations for slope and deflection are
The slope at B and x = 2 m, the third term is not valid
Deflection at C, x = 4 m, all the terms in equation for
deflection are valid for x = 4 m
4. A beam ABCD, 7 m long hinged at A and roller
supported at D carries 7 kN load at B and 4 kN/m udl
over BC = 3 m. If EI = 14,000 kN m2 for the beam,
determine the slope at A and deflection at point C.
Solution:
Reactions
Total udl on beam = 4 × 3 = 12 kN
CG of this load lies at 2 + 1.5 = 3.5 m from end A
Taking moments about A,
7 × 2 + 4 × 3(3.5) = 7 RD
RA = 7 + 12 − 8 = 11kN
Last portion of the beam is CD. A section YY at a
distance x from end A in the portion CD of the beam is
taken to make the equation of bending moment valid
for all the three portions. The udl is extended to
section YY on both sides (upward and downward), so
that its net effect becomes zero.
Bending moment at section YY
where w is the rate of loading.
Note that the first term is valid for portion AB, the first
three terms are valid for portion BC, and all the four
terms are valid for portion CD of the beam.
Substituting the value of w = 4 kN/m,
Integrating
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 3
where C1 is a constant of integration.
Also integrating (2) we obtain
where C2 is another constant of integration.
At end A, x = 0, y = 0 (in portion AB, x = 0, their three
terms are invalid)
Constant, C2 = 0.
At the end D, x = 7 m, y = 0, and all the terms in
equation are valid.
Constant, C1 = –54.5
The equations will become
EI = 14,000 kN m2
Slope at A, x = 0
Deflection at C, x = 5 m, and all the terms are valid
5. Cantilever AB, 5 m long, is simply supported
at A and fixed at B. If it carries a udl of 6 kN/m
over CB = 3 m, EI of cantilever is 3,600 kN m2.
Determine the reaction at A and slope at A and also find
out the deflection at C?
Solution:
Say reaction at A is RA. Taking a section at
distance x from A, in the portion CB,
Bending moment,
where w is rate of loading
or,
Integrating we get
where C1 is a constant of integration; dy/dx = 0 at x = 5,
by substituting this value, we obtain
So,
Also by integrating we obtain
At A, x = 0, y = 0
0 0 = − omitted term + C2
Constant, C2 = 0
Moreover, y = 0 at end B, x = 5 m, by substituting this
value,
RA = 2.754 kN
By substituting the value of RA, equations of slope and
deflection are
At A, x = 0
Deflection at C, x = 2 m
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 4
6. A simply supported beam of span length L carries a
udl of intensity w throughout its length. Determine the
slope at A and deflection at C by moment area method.
Solution:
BM diagram is parabolic for this case with maximum
bending moment at centre
Area of the bending moment diagram from A to C
that is, area AC′C
Because of symmetrical loading, ic = 0, and slope at
centre is zero.
CG of area AC’C lies at a distance of (5/8 × L/2) from
A.
Now,
7. A beam, 6 m long, is simply supported at ends
carries a udl of intensity 4 kN/m throughout its length.
Draw BM diagram of the beam. Using conjugate beam
method, determine the slope at the ends and deflection
in the centre. EI is the flexural rigidity of the beam. EI =
10,500 kN m2.
Solution:
BM diagram is parabolic with Mmax =
Bending moment diagram supported on two supports
is shown as conjugate beam.
Area of BM diagram
Slope at A,
Slope at B,
M′C = moment at centre
8. A beam of length L carries a central load W as
shown in figure. Moment of inertia for quarter length
from ends is I1 and for the middle half length moment
of inertia is I2, such that I2 = 2I1, now draw the
conjugate beam diagram.
Solution:
If E is Young’s modulus of the material, diagram AE′B is
the bending moment diagram such that
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 5
Conjugate beam diagram gives,
Beam is symmetrically loaded, therefore reactions, RA′
= RB′.
RA′ = area of conjugate beam diagram up to centre.
iE = slope at centre is zero
iE – iA = area of conjugate beam diagram up to centre.
Moment at centre, M′E
Deflection at centre =
9. A uniform beam of length L is supported
symmetrically over a span l. Determine the ratio
of l/L if the upward deflections at the ends is equal to
the downward deflection at the centre due to a
concentrated load at mid span.
Solution:
Fig. shows a beam AB of length L, supported
over CD = l, central load is W. Say EI is the flexural
rigidity of the beam. Reaction RC = RD = W/2 (on
account of symmetrical loading).
Let us analyze only half the portion of the beam and
slope at centre E will be zero. Take a section YY at a
distance of x from A, in portion CE.
Integrating
So,
Integrating
By substituting the value,
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 6
Deflection at A, yA, x = 0
Deflection at E, x = L/2
But yA = –yE (as they are numerically equal)
10. A propped cantilever of length L is fixed at one end
and roller supported at the other end. Cantilever is
subjected to a couple M at L/4 from A as shown in fig.
Determine reaction at propped end, deflection at C.
Solution:
The reaction at propped end is RA. Consider a
section YY at a distance x from A (on the portion CB).
Integrating
where C1 is a constant of integration.
Now dy/dx = 0 at x = L, so
So,
Also by integrating
C2 is another constant of integration,
0 = 0 – write then omitted term – 0 – 0 – C2
Now y = 0 at x = 0,
Constant, C2 = 0
Finally,
At fixed end, y = 0, x = L, by substituting this value
Deflection at C, x = L/4 by substituting
Substituting the value of RA
11. A long steel strip of uniform width and thickness
2.5 mm is lying on a level floor. Its one end is passing
over a roller of 60 mm in diameter lying on the floor at
one point. At what distance on either side of the roller
will the strip be clear of the ground? What is the
maximum bending moment in strip? ρ for steel = 76.44
× 10–6 N/mm3. Take b = 50 mm and E = 210 kN/mm2.
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 7
Solution:
b = 60 mm ; t = 2.5 mm;
w = b × t × ρ N/mm = weight per unit length
= 60 × 2.5 × 76.44 × 10−6 = 0.011466 N/mm.
Say L is the length of the strip which is clear from the
floor. At A, strip just leaves the ground. Say RC is
reaction at C and RA is reaction at A. Consider a
sections YY at a distance x from C.
Bending moment,
or
Integrating , we get
where C1 is a constant of integration.
0 = 0 – 0 + C1
Constant, C1 = 0
where w is rate of loading.
By integrating
where C2 is another constant of integration.
Now, x = L, y = 0, by substituting this value, we get,
Finally,
Moreover at end A, x = L, dy/dx = 0,
Substituting these values
by substituting the value of RC
At point C, y = –60 mm; x = 0. (we have taken x positive
towards right. The y is negative upwards).
Maximum bending moment occurs as a
E = 210 × 1,000 N.mm; w = 0.011466 N/mm
Substituting these values
L = 1894.2 mm = 1.894 m
12. A beam AB of length 4L is being lifted by a crane
and is bolted to the left and right hand cantilevers at A
and B. In raising this beam into position and before it is
properly aligned, it fouls at A and B carrying one
upward force P exerted by crane as shown in figure.
Show that the deflection of the beam at the crane hook
if E and I are same for each of the beam section is
1.328PL3/EI.
Solution:
Reaction at A, RA = 0.375 P
Reaction at B, RB = 0.625 P
Deflection at A,
Deflection at B,
δE due to P at E,
Total deflection at E
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 8
13. A beam AB is simply supported at ends of length 8
m, which carries a udl of 4 kN/m throughout its length.
It is propped at the centre by a force P such that the
level of the centre is the same as the level of ends.
Determine the magnitude of P. What are reactions at
support and bending moment at centre? If EI is the
flexural rigidity, what is the slope at ends?
Solution:
For a simply supported beam, span length L, carrying a
udl of intensity w, the central deflection is
For a simply supported beam, span length L with
central load P, the central deflection is
But net deflection at C is zero.
Substituting the values w = 4 kN/m, L = 8 m, we obtain
Propping force, P = 20 kN
Total load on the beam = 4 × 8 = 32 kN
Reaction, RA = RB is due to symmetric loading
RA + RB + P = 32 kN
RA + RB = 32 − 20 = 12 kN,
RA = RB = 6 kN
Bending moment at centre C
Taking moments about C, for portion AC
= -8 kN.m
Consider a section at a distance x from A,
y = 0 , x = 0, constant, C2 = 0
y = 0, x = 4 m
Slope at A, x = 0
Due to symmetry, slope at B,
14. A beam AB, 2 m long, carries a uniformly
distributed load of 10 kN/m is resting over a similar
beam (of same cross section) CD, 1 m long as shown
in Fig. Determine the reaction at C.
Solution:
Say the reaction at C = P′
Consider a section YY at a distance x from B,
Integrating
By integrating
y = 0 at x = 2 m
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 9
Now,
at x = 0 and deflection on CD
15. A beam of length L is hinged at both the ends. An
anticlockwise turning moment M is applied at point C.
Point C is at a distance of L/4 from end A. Determine
the slope and deflection at A. Given EI is the flexural
rigidity of the beam.
Solution:
For reactions, let us take moments about A
(e cw) M = RB × L (cw)
To balance RB, reaction
Consider the section YY at a distance of x from A in
portion CB of beam.
Bending moment,
or
By integrating
where C1 is constant of integration
Again integrating
at x = 0, y = 0, so
0 = 0 omitted term + C1 × 0 + C2
Constant, C2 = 0
At x = L, y = 0 at end B, by substituting this value
Expressions for slope and deflection are
at x = L/4, point C.
Slope,
Deflection
16. A simply supported beam of length L carries a
load W at point C, such that AC= a, CB = b, a + b = L.
Draw BM diagram for the beam and determine the
deflection under the load. Given EI as the flexural
rigidity of beam.
Solution:
BM diagram is shown with
Consider the portion AC, origin at A, x positive towards
right, first moment of area AC′C about origin A
Say slope at C is iC
Slope at A is iA.
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 10
Then,
or
Consider portion BC, x origin at B, x positive towards
left
By adding we obtain;
But iC = –iC, because slope is same and taking x positive
towards left, iC = –iC′
17. A simply supported beam of length L carries two
loads W each at L/4 from each end. Draw BM diagram
of beam and determine slope at ends and deflection at
centre. EI is the flexural rigidity of beam.
Solution:
Beam with load W at L/4 from each end is shown in fig.
Reaction, RA = RB = W (because of symmetrical loading).
Bending moment,
Slope at ends
Area of BM diagram up to middle of the beam abcd.
Area,
Because of symmetrical loading, iE = 0.
(because of symmetrical loading)
First moment of area about A
But,
So,
Deflection at centre,
18. A cantilever AB of length L is of different section
with moment of inertia I1for L/2 and I2 for L/2 to L as
shown in fig. Determine the slope and deflection at A
if E is Young’s modulus of the material.
Solution:
By integrating both the sides
At fixed end, iB = 0
Multiply the equation of bending moment
by xdx throughout
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 11
But slope iB = 0, deflection yB = 0 at fixed end.
So
19. A cantilever, 3 m long, carries a udl of 20 kN/m
throughout its length. Its free end is attached to a
vertical tie rod 2.4 m long and 20 mm in diameter. The
rod is initially straight. Determine the load taken by the
rod and deflection in cantilever at A. E = 210 kN/mm2
I = 1,000 × 104 mm4
Solution:
Tensile force P in rod produces reaction P in cantilever
as shown. Consider a section YY at a distance x for A.
Bending moment, Mx = Px – (wx2/2),
where w = 20 kN/m.
Integrating
where C1 is constant of integration.
dy/dx = 0 at B i.e., x = 3 m
By integrating
y = 0, at x = 3 m, end B
Deflection at A,
Extension in bar,
Change the length of the rod, dl = yA
Load taken by rod, P = 22.31 kN
Deflection in cantilever,
20. In a structure, a vertical member made of copper is
supported by two steel bars as shown in Fig. The
copper member is subjected to a temperature rise of
100°C. Calculate the maximum bending moment set up
in steel member which can be considered as simply
supported beam. Weight of all the members may be
neglected. Width of all the members is 20 mm.
For copper, EC = 105 kN/mm2, αC = 18 × 10–6/°C
For steel, ES = 200 kN/mm2
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 12
Solution:
Steel member
Free expansion of copper member is prevented by
= 300 × α × T – 2δ = 300 × 18 × 10–6 – 2 × 0.211 W
= 0.54 – 0.422 W
Area of cross section of copper member = 20 × 5
= 100 mm2
Compressive force in copper member,
21. A beam, 9 m long, is simply supported at ends
carries a linearly varying load of zero intensity at C
increasing to 12 kN/m at B over a distance of 6 m.
If EI is the flexural rigidity of the beam, determine the
slope and deflection at point C.
Solution:
Total load on beam = (6×12)/2 = 36 kN
CG of load from A, 3 + 4 = 7 m.
Moment about A, 36 × 7 = 9RB.
Reaction, RB = 28 kN
RA = 36 – 28 = 8 kN.
Take a section YY at a distance x from A
wx = rate of loading at x
Mx = bending moment at x
By integrating
x = 0, y = 0, C2 = 0
x = 9m, y = 0
C1 = -93.6
At C, x = 3
22. A composite beam consists of a wooden section,
240 mm width × 260 mm depth. Two steel plates of
width 240 mm and thickness 20 mm are fixed on the
top and bottom of wooden section as shown in Fig.
11.48. Composite beam is simply supported over a
span of 5 m and a load of 30 kN is applied at the centre
of the beam. Determine the deflection at the centre of
composite beam. ES = 200 GPa, EW = 15 GPa.
Figure 11.48
Solution:
Section is symmetrical about the central axis XX.
Moment of inertia,
Moment of inertia, IS
Mechanics of Solids Deflection Numericals
Yatin Kumar Singh Page 13

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Def numerical

  • 1. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 1 1. A steel girder of I section supported over a span of 3.6 m has Ixx = 1,200 × 104 mm4. Determine the slope at ends and deflection at the centre of the beam, if central load is 16 kN, E = 200 GPa = 200 × 109 N/m2 = 200 × 106 kN/m2. Solution: E = 200 × 106 kN/m2 I = 1,200 × 104 mm4 EI = 1,200 × 200 × 10-2 = 2,400 kNm2 W = 16 kN; L = 3.6m 2. A beam AB, 10 m long, carries point loads of 6 and 3 kN at C and D as shown in Fig. Determine support reactions, deflection at C and D, and slope at ends A and B, if EI is the flexural rigidity of the beam. Solution: Taking moments about A, 6 × 4 + 7 × 3 = 10RB Reaction, RB = (24 + 21)/10 = 4.5 kN Total load on beam = 6 + 3 = 9 kN Reaction at A, RA = 9 − RB = 9 − 4.5 = 4.5 kN. There are three portions, that is, AC, CD and DB in the beam and if A is the origin then DB is the last portion. Consider a section at a distance x from A in the portion DB, Bending moment, Mx = 4.5x − 6(x − 4) − 3(x − 7) Integrating Since the beam is not symmetrically loaded about its centre, so we do not know where slope is zero. By integrating eq (ii) we obtain, EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 + C1x + C2 Boundary conditions: x = 0 at end A, deflection y = 0 Moreover in portion AC, only the first term is valid and the other two terms of equation are not valid. So, 0 = 0.75 × 0, neglected term – neglected term + C1 × 0 + C2 or, constant C2 = 0 So, EIy = 0.75x3 – (x – 4)3 – 0.5(x – 7)3 + C1x At end B, x = 10 m, deflection y = 0, by substituting the value 0 = 0.75 × 103 – 63 – 0.5 × 33 + 10C1 0 = 750 – 216 –13.5 + 10C1 Constant, C1 = –52.05 Finally, the equations are EIy = 0.75x3 − (x − 4)3 − 0.5(x − 7)3 − 52.05x Slope at end A, x = 0 EIiA = 2.25 × 0 – neglected terms –52.05 at end B, x = 10, so all the terms are valid Deflection At point C, x = 4 m, hence, the third term in the equation is invalid. EIyC = 0.75 × 43 − (4 − 4)3 − neglected term − 52.05 × 4 EIyC = 48 − 0 − 208.2 = −160.2 At point D, x = 7 m, and all the terms in the equation for deflection are valid. EIyD = 0.75 × 73 − (7 − 4)3 − 0.5(7 − 7)3 − 52.05 × 7 = 257.25 − 27 − 0 − 364.35 = −134 1 3. A beam ABCD, 6 m long hinged at end A and roller supported at end D, is subjected to CCW moment of 10 kN m at point B and a point load of 10 kN at point C as shown in Fig. Determine the deflection under load of 10 kN and slope at point B, by taking EI as flexural rigidity of the beam. Solution: Taking moments at A, 10 + RD × 6 = 10 × 4 RD = 5kN ↑ Total load on the beam = 10 kN Reaction at A, RA = 10 – 5 = 5kN ↑
  • 2. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 2 There are three portions: AB, BC and CD in beam, with A as the origin portion and CD is the last. Take a section YY in portion CD at a distance x from A as shown in the figure. The equation of bending moment is Note that 10 kN m is a moment and we cannot take moment of moment, but 10 kN m is applied at B at a distance of (x – 2) m from section YY. Moreover (x – 2)° = 1, so only to locate the position of moment, the term (x – 2)° is taken, that is, (x – 2) raised to power zero, this term locates the position of moment applied at B. Integrating we get where C1 is a constant of integration, we do not know where the slope is zero as the beam is not symmetrically loaded. Also by integrating we obtain where C2 is another constant of integration. y = 0 at end A, x = 0, only first two terms in between are valid in portion AB, so or, constant, C2 = 0. y = 0 at end D, where x = 6 m, all the terms in the equation are valid, so Finally, the equations for slope and deflection are The slope at B and x = 2 m, the third term is not valid Deflection at C, x = 4 m, all the terms in equation for deflection are valid for x = 4 m 4. A beam ABCD, 7 m long hinged at A and roller supported at D carries 7 kN load at B and 4 kN/m udl over BC = 3 m. If EI = 14,000 kN m2 for the beam, determine the slope at A and deflection at point C. Solution: Reactions Total udl on beam = 4 × 3 = 12 kN CG of this load lies at 2 + 1.5 = 3.5 m from end A Taking moments about A, 7 × 2 + 4 × 3(3.5) = 7 RD RA = 7 + 12 − 8 = 11kN Last portion of the beam is CD. A section YY at a distance x from end A in the portion CD of the beam is taken to make the equation of bending moment valid for all the three portions. The udl is extended to section YY on both sides (upward and downward), so that its net effect becomes zero. Bending moment at section YY where w is the rate of loading. Note that the first term is valid for portion AB, the first three terms are valid for portion BC, and all the four terms are valid for portion CD of the beam. Substituting the value of w = 4 kN/m, Integrating
  • 3. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 3 where C1 is a constant of integration. Also integrating (2) we obtain where C2 is another constant of integration. At end A, x = 0, y = 0 (in portion AB, x = 0, their three terms are invalid) Constant, C2 = 0. At the end D, x = 7 m, y = 0, and all the terms in equation are valid. Constant, C1 = –54.5 The equations will become EI = 14,000 kN m2 Slope at A, x = 0 Deflection at C, x = 5 m, and all the terms are valid 5. Cantilever AB, 5 m long, is simply supported at A and fixed at B. If it carries a udl of 6 kN/m over CB = 3 m, EI of cantilever is 3,600 kN m2. Determine the reaction at A and slope at A and also find out the deflection at C? Solution: Say reaction at A is RA. Taking a section at distance x from A, in the portion CB, Bending moment, where w is rate of loading or, Integrating we get where C1 is a constant of integration; dy/dx = 0 at x = 5, by substituting this value, we obtain So, Also by integrating we obtain At A, x = 0, y = 0 0 0 = − omitted term + C2 Constant, C2 = 0 Moreover, y = 0 at end B, x = 5 m, by substituting this value, RA = 2.754 kN By substituting the value of RA, equations of slope and deflection are At A, x = 0 Deflection at C, x = 2 m
  • 4. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 4 6. A simply supported beam of span length L carries a udl of intensity w throughout its length. Determine the slope at A and deflection at C by moment area method. Solution: BM diagram is parabolic for this case with maximum bending moment at centre Area of the bending moment diagram from A to C that is, area AC′C Because of symmetrical loading, ic = 0, and slope at centre is zero. CG of area AC’C lies at a distance of (5/8 × L/2) from A. Now, 7. A beam, 6 m long, is simply supported at ends carries a udl of intensity 4 kN/m throughout its length. Draw BM diagram of the beam. Using conjugate beam method, determine the slope at the ends and deflection in the centre. EI is the flexural rigidity of the beam. EI = 10,500 kN m2. Solution: BM diagram is parabolic with Mmax = Bending moment diagram supported on two supports is shown as conjugate beam. Area of BM diagram Slope at A, Slope at B, M′C = moment at centre 8. A beam of length L carries a central load W as shown in figure. Moment of inertia for quarter length from ends is I1 and for the middle half length moment of inertia is I2, such that I2 = 2I1, now draw the conjugate beam diagram. Solution: If E is Young’s modulus of the material, diagram AE′B is the bending moment diagram such that
  • 5. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 5 Conjugate beam diagram gives, Beam is symmetrically loaded, therefore reactions, RA′ = RB′. RA′ = area of conjugate beam diagram up to centre. iE = slope at centre is zero iE – iA = area of conjugate beam diagram up to centre. Moment at centre, M′E Deflection at centre = 9. A uniform beam of length L is supported symmetrically over a span l. Determine the ratio of l/L if the upward deflections at the ends is equal to the downward deflection at the centre due to a concentrated load at mid span. Solution: Fig. shows a beam AB of length L, supported over CD = l, central load is W. Say EI is the flexural rigidity of the beam. Reaction RC = RD = W/2 (on account of symmetrical loading). Let us analyze only half the portion of the beam and slope at centre E will be zero. Take a section YY at a distance of x from A, in portion CE. Integrating So, Integrating By substituting the value,
  • 6. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 6 Deflection at A, yA, x = 0 Deflection at E, x = L/2 But yA = –yE (as they are numerically equal) 10. A propped cantilever of length L is fixed at one end and roller supported at the other end. Cantilever is subjected to a couple M at L/4 from A as shown in fig. Determine reaction at propped end, deflection at C. Solution: The reaction at propped end is RA. Consider a section YY at a distance x from A (on the portion CB). Integrating where C1 is a constant of integration. Now dy/dx = 0 at x = L, so So, Also by integrating C2 is another constant of integration, 0 = 0 – write then omitted term – 0 – 0 – C2 Now y = 0 at x = 0, Constant, C2 = 0 Finally, At fixed end, y = 0, x = L, by substituting this value Deflection at C, x = L/4 by substituting Substituting the value of RA 11. A long steel strip of uniform width and thickness 2.5 mm is lying on a level floor. Its one end is passing over a roller of 60 mm in diameter lying on the floor at one point. At what distance on either side of the roller will the strip be clear of the ground? What is the maximum bending moment in strip? ρ for steel = 76.44 × 10–6 N/mm3. Take b = 50 mm and E = 210 kN/mm2.
  • 7. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 7 Solution: b = 60 mm ; t = 2.5 mm; w = b × t × ρ N/mm = weight per unit length = 60 × 2.5 × 76.44 × 10−6 = 0.011466 N/mm. Say L is the length of the strip which is clear from the floor. At A, strip just leaves the ground. Say RC is reaction at C and RA is reaction at A. Consider a sections YY at a distance x from C. Bending moment, or Integrating , we get where C1 is a constant of integration. 0 = 0 – 0 + C1 Constant, C1 = 0 where w is rate of loading. By integrating where C2 is another constant of integration. Now, x = L, y = 0, by substituting this value, we get, Finally, Moreover at end A, x = L, dy/dx = 0, Substituting these values by substituting the value of RC At point C, y = –60 mm; x = 0. (we have taken x positive towards right. The y is negative upwards). Maximum bending moment occurs as a E = 210 × 1,000 N.mm; w = 0.011466 N/mm Substituting these values L = 1894.2 mm = 1.894 m 12. A beam AB of length 4L is being lifted by a crane and is bolted to the left and right hand cantilevers at A and B. In raising this beam into position and before it is properly aligned, it fouls at A and B carrying one upward force P exerted by crane as shown in figure. Show that the deflection of the beam at the crane hook if E and I are same for each of the beam section is 1.328PL3/EI. Solution: Reaction at A, RA = 0.375 P Reaction at B, RB = 0.625 P Deflection at A, Deflection at B, δE due to P at E, Total deflection at E
  • 8. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 8 13. A beam AB is simply supported at ends of length 8 m, which carries a udl of 4 kN/m throughout its length. It is propped at the centre by a force P such that the level of the centre is the same as the level of ends. Determine the magnitude of P. What are reactions at support and bending moment at centre? If EI is the flexural rigidity, what is the slope at ends? Solution: For a simply supported beam, span length L, carrying a udl of intensity w, the central deflection is For a simply supported beam, span length L with central load P, the central deflection is But net deflection at C is zero. Substituting the values w = 4 kN/m, L = 8 m, we obtain Propping force, P = 20 kN Total load on the beam = 4 × 8 = 32 kN Reaction, RA = RB is due to symmetric loading RA + RB + P = 32 kN RA + RB = 32 − 20 = 12 kN, RA = RB = 6 kN Bending moment at centre C Taking moments about C, for portion AC = -8 kN.m Consider a section at a distance x from A, y = 0 , x = 0, constant, C2 = 0 y = 0, x = 4 m Slope at A, x = 0 Due to symmetry, slope at B, 14. A beam AB, 2 m long, carries a uniformly distributed load of 10 kN/m is resting over a similar beam (of same cross section) CD, 1 m long as shown in Fig. Determine the reaction at C. Solution: Say the reaction at C = P′ Consider a section YY at a distance x from B, Integrating By integrating y = 0 at x = 2 m
  • 9. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 9 Now, at x = 0 and deflection on CD 15. A beam of length L is hinged at both the ends. An anticlockwise turning moment M is applied at point C. Point C is at a distance of L/4 from end A. Determine the slope and deflection at A. Given EI is the flexural rigidity of the beam. Solution: For reactions, let us take moments about A (e cw) M = RB × L (cw) To balance RB, reaction Consider the section YY at a distance of x from A in portion CB of beam. Bending moment, or By integrating where C1 is constant of integration Again integrating at x = 0, y = 0, so 0 = 0 omitted term + C1 × 0 + C2 Constant, C2 = 0 At x = L, y = 0 at end B, by substituting this value Expressions for slope and deflection are at x = L/4, point C. Slope, Deflection 16. A simply supported beam of length L carries a load W at point C, such that AC= a, CB = b, a + b = L. Draw BM diagram for the beam and determine the deflection under the load. Given EI as the flexural rigidity of beam. Solution: BM diagram is shown with Consider the portion AC, origin at A, x positive towards right, first moment of area AC′C about origin A Say slope at C is iC Slope at A is iA.
  • 10. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 10 Then, or Consider portion BC, x origin at B, x positive towards left By adding we obtain; But iC = –iC, because slope is same and taking x positive towards left, iC = –iC′ 17. A simply supported beam of length L carries two loads W each at L/4 from each end. Draw BM diagram of beam and determine slope at ends and deflection at centre. EI is the flexural rigidity of beam. Solution: Beam with load W at L/4 from each end is shown in fig. Reaction, RA = RB = W (because of symmetrical loading). Bending moment, Slope at ends Area of BM diagram up to middle of the beam abcd. Area, Because of symmetrical loading, iE = 0. (because of symmetrical loading) First moment of area about A But, So, Deflection at centre, 18. A cantilever AB of length L is of different section with moment of inertia I1for L/2 and I2 for L/2 to L as shown in fig. Determine the slope and deflection at A if E is Young’s modulus of the material. Solution: By integrating both the sides At fixed end, iB = 0 Multiply the equation of bending moment by xdx throughout
  • 11. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 11 But slope iB = 0, deflection yB = 0 at fixed end. So 19. A cantilever, 3 m long, carries a udl of 20 kN/m throughout its length. Its free end is attached to a vertical tie rod 2.4 m long and 20 mm in diameter. The rod is initially straight. Determine the load taken by the rod and deflection in cantilever at A. E = 210 kN/mm2 I = 1,000 × 104 mm4 Solution: Tensile force P in rod produces reaction P in cantilever as shown. Consider a section YY at a distance x for A. Bending moment, Mx = Px – (wx2/2), where w = 20 kN/m. Integrating where C1 is constant of integration. dy/dx = 0 at B i.e., x = 3 m By integrating y = 0, at x = 3 m, end B Deflection at A, Extension in bar, Change the length of the rod, dl = yA Load taken by rod, P = 22.31 kN Deflection in cantilever, 20. In a structure, a vertical member made of copper is supported by two steel bars as shown in Fig. The copper member is subjected to a temperature rise of 100°C. Calculate the maximum bending moment set up in steel member which can be considered as simply supported beam. Weight of all the members may be neglected. Width of all the members is 20 mm. For copper, EC = 105 kN/mm2, αC = 18 × 10–6/°C For steel, ES = 200 kN/mm2
  • 12. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 12 Solution: Steel member Free expansion of copper member is prevented by = 300 × α × T – 2δ = 300 × 18 × 10–6 – 2 × 0.211 W = 0.54 – 0.422 W Area of cross section of copper member = 20 × 5 = 100 mm2 Compressive force in copper member, 21. A beam, 9 m long, is simply supported at ends carries a linearly varying load of zero intensity at C increasing to 12 kN/m at B over a distance of 6 m. If EI is the flexural rigidity of the beam, determine the slope and deflection at point C. Solution: Total load on beam = (6×12)/2 = 36 kN CG of load from A, 3 + 4 = 7 m. Moment about A, 36 × 7 = 9RB. Reaction, RB = 28 kN RA = 36 – 28 = 8 kN. Take a section YY at a distance x from A wx = rate of loading at x Mx = bending moment at x By integrating x = 0, y = 0, C2 = 0 x = 9m, y = 0 C1 = -93.6 At C, x = 3 22. A composite beam consists of a wooden section, 240 mm width × 260 mm depth. Two steel plates of width 240 mm and thickness 20 mm are fixed on the top and bottom of wooden section as shown in Fig. 11.48. Composite beam is simply supported over a span of 5 m and a load of 30 kN is applied at the centre of the beam. Determine the deflection at the centre of composite beam. ES = 200 GPa, EW = 15 GPa. Figure 11.48 Solution: Section is symmetrical about the central axis XX. Moment of inertia, Moment of inertia, IS
  • 13. Mechanics of Solids Deflection Numericals Yatin Kumar Singh Page 13