Mechanics of structures module2

Mechanics of solidsMechanics of solids
Torsion,
Bending moment and shear forceBending moment and shear force
Dr. Rajesh K. N.
Assistant Professor in Civil EngineeringAssistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
1

Module IIModule II
Torsion - torsion of circular elastic bars - statically
indeterminate problems - torsion of inelastic circular bars
Axial force, shear force and bending moment -
diagrammatic conventions for supports and loading, axialdiagrammatic conventions for supports and loading, axial
force, shear force and bending moment diagrams - shear
force and bending moments by integration and by
singularity functions
Dept. of CE, GCE Kannur Dr.RajeshKN
2

TorsionTorsion
Torsional moment (Torque)
Twist (Angle of twist)
• Circular shafts and non-circular shafts
3

Assumptions in torsion theory for circular shaftsAssumptions in torsion theory for circular shafts
• Material is uniform throughout
• Shaft remains circular after loading
• Plane sections remain plain after loading
T i i if h h• Twist is uniform throughout
• Distance between any two normal sections remain the same after loading
• Stresses are within elastic limit• Stresses are within elastic limit
4

T
ϕ θ
m’
r
T
ϕ θ
m
O
r
T
L
Any cross-section
Any radial distance r
mm L rφ θ′ = =
Angle of twist per unit length
L
θ
→
mm L rφ θ
G
τ
φ =Shear strain
L
θ
O
r
rτ θ
∴ =
G
rθ
φ =But
5
G L
∴
L
φ =But

T
Torsion equation
R
T
θ
r
τ
θ
L
T
O
maxG
r L R
τ θ τ
∴ = =
r
G L
τ θ
=
Aδ
maxr
R
τ
τ⇒ =
. .T A rδ τ δ=
2max max
.
r
T A r r A
R R
τ τ
δ δ δ= =
Torsional moment on the elemental area =
2max
T T r A
R
τ
δ δ= =∫ ∫
R R
Total torsional moment on the section,
A A
R∫ ∫
2max
r A
R
τ
δ= ∫ max
J
R
τ
=
,
6
A
R R
Polar moment of inertiaJ →

T τ T GθmaxT
J R
τ
= T G
J r L
τ θ
= =
T
GJ L
θ
= GJ Torsional rigidity
GJ L
maxT Tτ J R Torsional section modulusmax
max
J R J R
τ= ⇒ = J R Torsional section modulus
4
dπ
ZZ XX YYJ I I I= = +Polar moment of inertia
32
dπ
=
7

Power transmittedo e t a s tted
W T θ=
Work done (per second) by torque T making a twist θ1 /second
1.W T θ=
1.P T θ=i.e., Power transmitted
2
60
nT
P
π
=If n is the rotation per minute,
k
1 Watt 1 Joule/Second = 1 Nm/s= =
1 HP 0.75 kW=
8

Problem 1: A hollow shaft is to transmit a power of 300 kW at 80 rpm.
If the shear stress is not to exceed 60 MN/m2 and internal diameter is
0.6 of external diameter, find the internal and external diameters
assuming that the maximum torque is 1.4 times the mean torque.g q q
300 kWP =
2
60
nT
P
π
=80 rpmn =
60
2
mean
P
T T
nπ
∴ = =
3
60 300 10
35809.862 Nm 35.81 kNm
2 80
meanT
π
× ×
= = =
×
max 1.4 meanT T∴ = 1.4 35.81 50.134 kNm= × =
maxT
J R
τ
= ( ) ( )( )44 4 4
0.6
32 32
J D d D D
π π
= − = − ( )
4
0.8704
32
Dπ
=
9
J R 32 32

3 6
50 134 10 60 10× ×
( )
4
50.134 10 60 10
2
0.8704
32
D Dπ
× ×
=
3
0.00488914D = 0.169 mD⇒ =
0.6 0.102 md D⇒ = =0.6 0.102 md D⇒
10

Problem 2: A solid circular shaft is to transmit 75 kW power at 200 rpm.
If the shear stress is not to exceed 50 MPa, and the twist is not to exceed
10 in 2 m length of shaft, find the diameter of shaft. G = 100 GPa.
75 kWP =
2
60
nT
P
π
=200 rpmn =
3
60 60 75 10
3581 Nm
2 2 200
P
T
× ×
= = =
×2 2 200nπ π ×
maxT Gτ θ 4
D
J
π 2
50 N mmτ =max
J R L
= =
32
J = max 50 N mmτ =
6
4
3581 50 10
2
32
DDπ
×
=
⎛ ⎞
⎜ ⎟
⎝ ⎠
0.0714 mD⇒ =maxT
J R
τ
= ⇒
11
32⎝ ⎠

T G
J L
θ
=
9
4
3581 100 10 1 180
2
32
D
π
π
× × ×
⇒ =
⎛ ⎞
⎜ ⎟
⎝ ⎠
0.0804 mD⇒ =
32⎝ ⎠
Required diameter of shaft is the greater of the two values
0.0804 m 80.4 mmD∴ = =
12

Problem 3: A steel bar of 20 mm diameter and 450 mm length fails at a
torque of 800 Nm. What is the modulus of rupture of this steel in
torsion?
Modulus of rupture in torsion is the maximum shear
stress (on the surface) at failure.
T
R J
τ
=
Hence, modulus of rupture
TR
J
τ = where T is the torque at failure.
3
2800 10 10
509N
× ×
J
2
4
800 10 10
509N mm
20
32
τ
π
= =
⎛ ⎞×
⎜ ⎟
⎝ ⎠

Problem 4: A solid brass rod AB (G=39 Gpa, 30 mm dia, 250 mm length) is
b d d lid l i i d BC (G 27 G 36 di 320 l h)bonded to solid aluminium rod BC (G=27 Gpa, 36 mm dia, 320 mm length).
Determine the angles of twist at A and B.
T G TL
J L GJ
θ
θ= ⇒ =
A
BC
3
3
180 10 320
27 10 164895.92
Bθ
× ×
=
× ×
0
0.0129 rad 0.74= =
Torque, T
180 Nm
4
436
164895.92mm
32
BCJ
π ×
= =
3
3
180 10 250
39 10 79521 56
ABθ
× ×
=
× ×
0
0.0145 rad 0.831= =
4
430
79521.56mm
32
ABJ
π ×
= =
39 10 79521.56× ×
0 00
0.74 0.83 . 711 1 5ACθ = + =

Problem 5: Two solid brass rods (G=39 Gpa) AB (30mm dia, 1.2m length) & BC
(40 di 1 8 l h) li d i h h D i h(40mm dia, 1.8m length), are applied with torques as shown. Determine the
angles of twist between i) A and B; ii) A and C.
T G TL
J L GJ
θ
θ= ⇒ =
3
3
400 10 1200
39 10 79521.56
BAθ
× ×
=
× ×
0.1548 rad=
3
3
800 10 1800
39 10 251327.41
CBθ
− × ×
=
× ×
0.1469 rad= −
4
430
79521.56mm
32
ABJ
π ×
= =θ θ θ= +
4
440
251327.41 mmBCJ
π ×
= =
32
AB
CA BA CBθ θ θ= +
0.1548 0.1469 0.0079 radCAθ = − =
251327.41 mm
32
BCJ

Statically indeterminate shaftsy
A
B
T T
B
C
Torque, T
A BT T T= + (1)
A
B
0Aθ = with respect to B
0θ θ− =
Torque, T
B
C
0CB ACθ θ =
0B BC A CAT L T L
G J G J
⇒ − =Torque, T
Reactive
torque, TA
Reactive
torque, TB
BC BC CA CAG J G J
(2)
Solve (1) and (2) for &A BT T

Torsion of inelastic circular bars
• Linear elastic range: Stress strain curve is a straight line
(Torsion of circular bars beyond elastic range)
g g
Stress
Strain
• Corresponding shear stress distribution in the shaft:
τ
T
r
J
τ =
maxτ
R Entire cross-section is in the
elastic range

Materials in the inelastic ranges
• Inelastic: Nonlinear stress strain curve
g
Stress
Elastic rebound
on unloading
• Elastic plastic : Initially in the elastic range then fully plastic
Strain
• Elastic-plastic : Initially in the elastic range, then fully plastic
Stress
Elastic rebound
on unloading
S
Strain

Shear stress-strain curve Corresponding shear stress
distribution in the shaft
ss
maxτStres
Strain
R
ss
maxτ
Stres
Strain
R

maxτ maxτFor any stress distribution,
. .dT dA rτ=
T dT r dAτ= =∫ ∫
R R
y ,
maxτ
R
.
A A
T dT r dAτ∫ ∫

Problem 1:
A solid steel shaft of 24 mm diameter is so severely twisted that only an 8 mm
diameter inner core remains elastic, while the rest of the diameter goes to
inelastic range. If the material is elastic-plastic with shear stress-strain diagram
h h h id l d id l i h ill i has shown, what are the residual stress and residual twist that will remain at the
surface? Take G = 80 GPa.
τ
160 MPa
4 mm
12 mm
0.002 γ φ=

The inner core is in the elastic rangeThe inner core is in the elastic range. 160 MPa
12 mm4 mm
It is required to find the applied torque.
∫
R R
∫ ∫
Applied torque,
. .
A
T r dAτ′ = ∫
2
0 0
. .2 . .2 .r r dr r drτ π τ π= =∫ ∫
4 12
160⎛ ⎞
( )
4 12
2 2
0 4
160
.2 . 160 .2 .
4
r r dr r drπ π⎛ ⎞
= +⎜ ⎟
⎝ ⎠
∫ ∫
( ) 3 3
16 558 10 Nmm 574 10 Nmm= + × = ×

Residual stresses on rebound (removal of torque)
• Consider the external torque is removed after a portion of the cross-section
has gone to inelastic range
( q )
has gone to inelastic range.
• After the torque has been removed completely, the inner portion has a
tendency to rotate back to original position but the outer portion which has atendency to rotate back to original position, but the outer portion, which has a
permanent rotation, prevents this.
• Outer portion has a tendency to stay in the permanently set position, butOuter portion has a tendency to stay in the permanently set position, but
the inner portion, which has a tendency to rotate back to original position,
prevents this.
• Hence, stresses are remaining in the shaft even after the removal of torsion.
• The rebound is elastic.
• If the deformation of the outer portion was elastic, shear stress would have
reached a value of τ’ at the surface.

• The stress recovery at the surface is τmax .
• Hence, as a result of inner portion applying a torsion on the outer portion, a
stress remains at the surface, equal to τ’ - τmax , opposite in direction to the
li d t i it t th di ti fapplied torque. i.e., opposite to the direction of τmax .
• Similarly, all the inner regions at various radial distances have residual
stressesstresses.
• These residual stresses are obtained as the difference between elastic-plastic
stress distribution and the elastic stress distribution of rebound as shown instress distribution and the elastic stress distribution of rebound, as shown in
figure.
Elastic rebound
τ′E g : Elastic plastic material
Residual stress
maxτ
E.g.: Elastic-plastic material

To find residual stresses on an elastic rebound, (i.e., on removal of the
T R′ 3
2574 10 12× ×
torque)
Shear stress on the surface T R
J
τ′ = 2
4
574 10 12
211 N mm
24
32
π
× ×
= =
⎛ ⎞×
⎜ ⎟
⎝ ⎠
(considering elastic rebound)
Hence, residual shear stress on the surface
2
211 160 51 N mm= − =
maxτ τ′= −
160 MPa
Residual
stress
211 MPa
12 mm4 mm

0.006γ φ′ ′= =
Residual twist on rebound
Strain variation is linear along the radius. 4 0.002φ =
Residual twist on rebound
θ
12mmR =
4mmmax max
L RG
θ τ
=At the initiation of yield,
maxT T=
When T increases further to T’,
L
θ
increases beyond max
, ot
L L
θ θ′ Strain variation
L L L
L rG
θ τ
=After yielding , is invalid from r=4 to 12.Note:
Final twist
(after yielding,
when T=T’)
θ′
i.e., .
L RG
θ τ′ ′
≠ But, r
L r R
θ φ φ′ ′
= =
On elastic rebound, recovered twist Elastic Rebound
(recovery of twist)
T
L GJ
θ ′
=
Twist of a radial line
(recovery of twist)

To find residual twist on an elastic rebound,
3
0.002
4 10
0.5 rad
m
m−
= =
×
r
L r
θ φ′
=
Final twist per unit length (after yielding,
when T=T’)
L r
3
rad m
160
4
m 0.5 ra
10
d
8
m
0
= =
× ×
or,
L rG
θ τ′
=
4 1080L rG
L rG
θ τ
=∵ is valid from r=0 to 4.
Elastic twist recovered (considering elastic rebound)
4
9 2 12 4
574Nm
24
80 10 N m 10 m
π −
=
⎛ ⎞×
× × ×⎜ ⎟
( g )
T
L GJ
θ ′
= 0.22 rad m=
Hence, residual twist 0.5 0.22 0.28rad m= − =
80 10 N m 10 m
32
× × ×⎜ ⎟
⎝ ⎠

Problem 2:
A solid circular shaft 1 2 m long and 50 mm diameter is subjected to a torque ofA solid circular shaft, 1.2 m long and 50 mm diameter is subjected to a torque of
4.6 kNm. Assuming the shaft is made of an elastoplastic material with yield
strength in shear of 150MPa and G = 80 Gpa, what are the radius of the elastic
core and angle of twist of the shaft? Also, what are the residual stresses andcore and angle of twist of the shaft? Also, what are the residual stresses and
residual (permanent) angle of twist of the shaft, after removal of torque?
A li d t
150 MPa
25 mmρ
To find elastic core.
. .
A
T r dAτ′ = ∫
2
0 0
. .2 . .2 .
R R
r r dr r drτ π τ π= =∫ ∫
Applied torque, 25 mmρ
( )
25
6 2 2
0
150
4.6 10 N.mm .2 . 150 .2 .r r dr r dr
ρ
ρ
π π
ρ
⎛ ⎞
× = +⎜ ⎟
⎝ ⎠
∫ ∫
( )3 33
6
2 252
4.6 10 N.mm 150 150
4 3
π ρπρ −
× = +
28
15.743 mmρ =∴

To find angle of twist
γ ′
To find angle of twist.
(Final angle of twist in the inelastic (plastic) range) ?rφ =
25 mmR =15.743 mmρ =
G
L
τ θ
ρ
= (Valid from r = 0 to 15.743 only)
Strain variation along radius
3
150 1200
80 10 15.743
L
G
τ
θ
ρ
×
∴ = =
× ×ρ
3
142.92 10 rad−
= ×
Final twist
(after yielding,
when T=T’)
θ′
0
3 0180
142.9 8. 82 0 11
π
−
= × × =
Elastic Rebound
(reco er of t ist)
29Twist of a radial line
(recovery of twist)

T R′
6
4.6 10 25× ×
torque)
Shear stress on the surface 2
187 52NT R
J
τ′ = 4
50
32
π
=
⎛ ⎞×
⎜ ⎟
⎝ ⎠
2
187.52N mm=
2
187.52 150 37.52 N mm= − =
maxτ τ′= −
150 MPa
Residual
stress
187.52 MPa
25 mm15.743 mm

3
4
9 2 12 4
4.6 10 Nm 1.2m
50
80 10 N m 10 m
π −
× ×
=
⎛ ⎞×
× × ×⎜ ⎟
T L
GJ
θ
′
=
0
0.1124 rad 6.44= =
Hence residual twist 0
8 18 6 4 74 1 4= − =
80 10 N m 10 m
32
× × ×⎜ ⎟
⎝ ⎠
Hence, residual twist 8.18 6.4 74 1. 4= =
Final twist
(after yielding,
h T=T’)
θ′
when T=T )
Elastic Rebound
(recovery of twist)
Twist of a radial line
(recovery of twist)

Problem 3:
A solid circular steel shaft 0 6 m long and 32 mm diameter has been twistedA solid circular steel shaft 0.6 m long and 32 mm diameter has been twisted
through 60. Steel is elastoplastic with yield strength in shear of 145MPa and G
= 77 Gpa. What is the maximum residual stress in the shaft, after removal of
torque?torque?
To find elastic core.
6 0.1047 rad
180
π
θ = × = 145 MPa
G
L
τ θ
ρ
= (Valid from r = 0 to ρ only)
16 mmρ
10.792 mm=
ρ
3
145 600
10.792 mm
77 10 0.1047
L
G
τ
ρ
θ
×
∴ = = =
× ×
32

Applied torque
To find torque applied.
. .
A
T r dAτ′ = ∫
2
0 0
. .2 . .2 .
R R
r r dr r drτ π τ π= =∫ ∫
Applied torque,
A 0 0
( )
10.792 16
2 2
0 10 792
145
.2 . 145 .2 .
10.792
r r dr r drπ π⎛ ⎞
= +⎜ ⎟
⎝ ⎠
∫ ∫0 10.792⎝ ⎠
( )3 33 2 16 10.7922 10.792
145 145
4 3
ππ −
= +
1148475.884 NmmT′∴ =
4 3

T R′
′
1148475.884 Nmm 16×
torque)
Shear stress on the surface
2T R
J
τ′ = 4
1148475.884 Nmm 16
32
32
π
=
⎛ ⎞×
⎜ ⎟
⎝ ⎠
2
178.5N mm=
2
178.5 145 33.5 N mm= − =
maxτ τ′= −
max
TR
J
τ= −Residual shear stress at 10.792 mmρ =
145 MPa
Residual
stress
178.5 MPa
max4
1148475.884 Nmm 10.792
32
τ
π
×
= −
⎛ ⎞×
⎜ ⎟
16 mm10.792 mm
2
120.4 145 24.6 N mm= − = −
32⎜ ⎟
⎝ ⎠
Hence the maximum residual shear stress is on the surface.

Problem 4:
A hollow circular steel shaft 1 25 m long 60 mm outer diameter and 36 mmA hollow circular steel shaft 1.25 m long, 60 mm outer diameter and 36 mm
inner diameter has been applied with a torque such that the inner surface first
reaches plastic zone, and then torque is removed. Steel is elastoplastic with
yield strength in shear of 145MPa and G = 77 Gpa What is the maximumyield strength in shear of 145MPa and G 77 Gpa. What is the maximum
residual stress in the shaft and residual (permanent) angle of twist, after
removal of torque?
30
Applied torque,
To find torque applied.
. .
A
T r dAτ′ = ∫ 2
0 0
. .2 . .2 .
R R
r r dr r drτ π τ π= =∫ ∫ ( )
30
2
18
145 .2 .r drπ= ∫
( )( )3 3
2 30 18
145
3
π −
=
145 MPa
6428452.551 NmmT′∴ =
( )4 4
60 36π × −
18 mm 30 mm
35
( ) 4
60 36
1107449.11 mm
32
J
π ×
= =

T R
J
τ
′
′ =
6428452.551 Nmm 30
1107449 11
×
=
torque)
Shear stress on the surface
(considering elastic rebound) J 1107449.11(considering elastic rebound)
2
174.142 N mm=
Hence, residual shear stress on the outer surface
2
174.142 145 29.142 N mm= − =
maxτ τ′= −
τ′
max
TR
J
τ= −Residual shear stress at 18 mmρ =
174 142 MPa=
145 MPa
Residual
stress
τ
max
6428452.551 Nmm 18
1107449.11
τ
×
= −
174.142 MPa
18 mm 30 mm 2
104.49 145 40.51 N mm= − = −
Hence the maximum residual shear stress is on the inner surface.

Final twist per unit length (after
yielding, when T=T’)
L
rG
τ
θ′ =
L rG
θ τ
=∵ is valid from r=0 to 18.
3
145 1250
rad 0.131 rad
18 77 10
×
= =
× ×
8 77 0
3 2 4
6428452.551 Nmm 1250mm
77 10 N mm 1107449 11mm
×
=
× ×
T L
GJ
θ
′
= 0.094 rad=
Hence, residual twist 0.131 0.09 0.034 7rad= − =
77 10 N mm 1107449.11mm× ×GJ
0
0.037rad 2.1= =

Bending moment and shear force
T f l d
g
Types of loads
Point load
Distributed loadDistributed load
Uniformly distributed load
Uniformly varying loady y g
Couple
T f tTypes of supports
Fixed (built-in or encastre)
Hinged (pinned)Hinged (pinned)
Roller
Guided fixed
38
Elastic (Spring)

Beams
Statically determinate and indeterminate beams
Beams
y
• Simply supportedSimply supported
• Cantilever
• Propped cantilever
• Fixed
• Continuous
39

Sign conventions
Bending Momentg
S i H iSagging Hogging
Shear Force
Clockwise Anticlockwise
40
Clockwise Anticlockwise

Shear force and bending moment diagramsShear force and bending moment diagrams
1 Cantilever1. Cantilever
• with a single load at the free end
• with several point loadswith several point loads
• with UD load over full span
• with UD load over part spanp p
• with a couple
• with combination of loads
41

B
l
B
x
(+)P
SFD
(-)
Pl−
42
BMD

l
B
x
wl (+)
SFDSFD
2
wl−
(-)
43
BMD

Si l t d b
• with a single point load at the centre
Simply supported beams
• with a single point load at the centre
• with a single eccentric point load
• with several point loadsp
• with UD load over full span
• with UD load over part span
h b f l d• with combination of loads
• with a couple
44

C
P
( ) 2
SFD
C
A B
2
P−
(+)
(-)SFD 2
4
PL
A B
45
BMD C

2
wL
(+)
2
wL−
SF diagram
CA
B
2 (+)
(-)
SF diagram
2
8
wL
A B
(+)
2
8
wL
A B
(+)
46BM diagram
C
BM diagram
C

P
ba
x
Pb
( )L
SFD
CA
B
Pa
L
−
(+)
(-)
SFD L
Pab
LA
B
(+)
BMD C

60 kN 50 kN
1
A D C B
1m
3m
6m
x
48

Simply supported beam with overhangp y pp g
• Overhang on one side with point loadOverhang on one side with point load
• Overhang on one side with UD load over full span
• Overhang on both sides with point loadsg
• Overhang on both sides with UD load over full span
49

P
ba
x
60 kN
30 kN/m
5 kN/m25 kNm
A DCB E F
1m 2m1m 1.5m 2.5m
50

Relation between shear force and bending momentRelation between shear force and bending moment
w
M M dM+
A C
V
V dV+
B D
0Y =∑
0BM =∑
dx
0Y =∑
( ) . 0V V dV w dx− + − = ( ) ( )
2
0
2
dx
M w V dV dx M dM+ + + − + =
.dV w dx− =
dV−
0Vdx dM− =
dM
V =
51
w
dx
= V
dx
=

Example 1Example 1
2
x
Px
M =
2
x
x
dM P
V
dx
= = Shear force
0xdV
dx
−
= Load intensity

Example 2Example 2
w kN/m
Lx L
2
wL
2
wL
2
2 2
x
wLx wx
M = −
2
2
x
x
dM wL
wx V
dx
= − = Shear force
xdV
w
dx
−
= Load intensity
dx

Example 3Example 3
l
B
l
x
x is taken in the negative direction
2
2
x
wx
M
−
=
x is taken in the negative direction
Hence dx is negative
2
x x
x
dM dM
wx V
dx dx
−
= = = Shear forcedx dx−
x xdV dV
w
dx dx
−
= = Load intensity
Shear force
54
dx dx−
y

Shear force and bending moment diagrams by integrationShear force and bending moment diagrams by integration
dV
w
−
=
dM
V =( ) 1V wdx C∴ = − +∫ 2M Vdx C∴ = +∫
• Slope of SFD at any point is the load intensity at that cross section
dx dx
( ) 1∫ 2∫
Slope of SFD at any point is the load intensity at that cross section
• Slope of BMD at any point is the shear force at that cross section
• Area of load diagram on the left (or right) of a cross section +g ( g )
reaction = shear force at that cross section
• Area of SFD on a segment of a beam = Change in bending moment
at that cross section i.e., dM Vdx=
• For bending moment M to be a maximum or minimum,
i.e.0 0,
dM
V= =
i.e.0 0, V
dx

Example 1
Total load = W
maxw L
W=
max
2
2
W
W
w
L
=
∴ =
L
L
3
W 2
3
W
dV−
( )∫
2w W
To draw SFD
dV
w
dx
= ( ) 1V wdx C∴ = − +∫ max
2
2
At ,
w W
x w x x
L L
= =
2
1 12 2
2 2
2
W W x
V xdx C C
L L
− −
∴ = + = +∫

22
12
Wx
V C
L
−
∴ = +
W
W W
At 0,
3
W
x V= =
x
maxw W
1 10
3 3
W W
C C∴ = + ⇒ =
max
2
2
w x x
L L
= =2
2
3
Wx W
V
L
∴
−
+= Parabolic variation
W
3 2
3
W−
SFD
3SFD

To draw BMD
dM
V
dx
=
2
22
3
Wx W
M dx C
L
⎛ ⎞−
∴ = + +⎜ ⎟
⎝ ⎠
∫⎝ ⎠
3
22
i.e.,
3 3
Wx Wx
M C
L
−
= + +
3 3L
At 0 0x M= = 2i.e., 0 C=
3
Wx Wx
M
−
∴ = +At 0, 0x M 2,
2
3 3
M
L
∴ = +
Cubic variation
i.e.0 0,
dM
V= =
For maximum bending moment,
2
9 3
WL
i.e.0 0, V
dx
2
0
Wx W L−
+ ⇒
9 3
3L
BMD
2
0
3 3
x
L
+ = ⇒ =

Example 2
d
( ) 1V wdx C= − +∫
PP
To draw SFD
∫
F t0 0
L
Portion AB
L/4 L/4
A
B C
D
From to0 , 0
4
x x w= = =
V C∴ =
L
PP
Constant1V C∴ =
But at 0,x V P= = 1C P⇒ = from 0 to,
L
x xV P =∴ ==
Constant
, 1 from 0 to
4
, x xV P∴
3L L L
Portion BC
From to
3
, 0
4 4
L L
x x w= = = Also, at , 0
4
L
x V= =
3L L
3
from to
4
0,
4
L L
xV x= =∴ = Constant

P
P−P−
SFD
To draw BMD
Portion AB
2M Vdx C= +∫
2M Pdx C∴ = +∫ 2Px C= +From to0 ,
4
L
x x V P= = =
Also, at 0, 0x M= = 2 0C⇒ = , from 0
4
to
L
M Px x x∴ = = =
Linear variation

Portion BC
2M Vdx C= +∫
Constant2M C∴ =From to
3
, 0
4 4
L L
x x V= = =
,
4 4
Also, at
L PL
x M= = 2
4
PL
C⇒ = from
3
,
4
t
4
o
4
PL L L
M x x∴ = = =
4 4 4
L L
4
PL
4
PL
BMD4 4

SummarySummary
Torsion - torsion of circular elastic bars - statically indeterminate
problems - torsion of inelastic circular bars
Axial force, shear force and bending moment - diagrammatic
conventions for supports and loading, axial force, shear force and
bending moment diagrams - shear force and bending moments bybending moment diagrams shear force and bending moments by
integration and by singularity functions
62

Mechanics of structures module2

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