mechanics of structures module2
DESCRIPTION
GCE KannurTRANSCRIPT
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 EngineeringGovt. College of Engineering, Kannur
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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, axial diagrammatic 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
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TorsionTorsionTorsional moment (Torque)
Twist (Angle of twist)
• Circular shafts and non-circular shafts
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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
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T
ϕ θ
m’r
T
ϕ θ
mO
r
TL
Any cross-sectionAny radial distance r
mm L rφ θ′ = =
Angle of twist per unit lengthLθ→
mm L rφ θ
Gτφ =Shear strain
L
θO
r
rτ θ∴ =
G
rθφ =But
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G L∴
Lφ =But
T
Torsion equation
R
Tθ
rτ
θ
LT O
maxGr L Rτ θ τ
∴ = =r
G Lτ θ=
AδmaxrR
ττ⇒ =
. .T A rδ τ δ=
2max max.rT A r r AR R
τ τδ δ δ= =
Torsional moment on the elemental area =
2maxT T r AR
τδ δ= =∫ ∫
R R
Total torsional moment on the section,A A R∫ ∫
2max r AR
τ δ= ∫ max JR
τ=
,
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AR R
Polar moment of inertiaJ →
T τ T GθmaxTJ R
τ= T G
J r Lτ θ
= =
TGJ L
θ= GJ Torsional rigidity
GJ L
maxT Tτ J R Torsional section modulusmaxmaxJ R J R
τ= ⇒ = J R Torsional section modulus
4dπZZ XX YYJ I I I= = +Polar moment of inertia
3 2dπ
=
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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
260nTP π
=If n is the rotation per minute,
k
1 Watt 1 Joule/Second = 1 Nm/s= =
1 HP 0.75 kW=
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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 is0.6 of external diameter, find the internal and external diametersassuming that the maximum torque is 1.4 times the mean torque.g q q
300 kWP = 260nTP π
=80 rpmn =
602mean
PT Tnπ
∴ = =
360 300 10 35809.862 Nm 35.81 kNm2 80meanTπ
× ×= = =
×
max 1.4 meanT T∴ = 1.4 35.81 50.134 kNm= × =
maxTJ R
τ= ( ) ( )( )44 4 4 0.6
32 32J D d D Dπ π= − = − ( )
4
0.870432Dπ
=
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J R 32 32
3 650 134 10 60 10× ×
( )4
50.134 10 60 1020.8704
32D Dπ
× ×=
3 0.00488914D = 0.169 mD⇒ =
0.6 0.102 md D⇒ = =0.6 0.102 md D⇒
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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 exceed10 in 2 m length of shaft, find the diameter of shaft. G = 100 GPa.
75 kWP = 260nTP π
=200 rpmn =
360 60 75 10 3581 Nm2 2 200
PT × ×= = =
×2 2 200nπ π ×
maxT Gτ θ 4DJ π 250 N mmτ =max
J R L= =
32J = max 50 N mmτ =
6
4
3581 50 102
32DDπ×
=⎛ ⎞⎜ ⎟⎝ ⎠
0.0714 mD⇒ =maxTJ R
τ= ⇒
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32⎝ ⎠
T GJ L
θ=
9
4
3581 100 10 1 1802
32D
ππ
× × ×⇒ =
⎛ ⎞⎜ ⎟⎝ ⎠
0.0804 mD⇒ =
32⎝ ⎠
Required diameter of shaft is the greater of the two values
0.0804 m 80.4 mmD∴ = =
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Problem 3: A steel bar of 20 mm diameter and 450 mm length fails at atorque of 800 Nm. What is the modulus of rupture of this steel intorsion?
Modulus of rupture in torsion is the maximum shear stress (on the surface) at failure.
TR Jτ=
Hence, modulus of ruptureTRJ
τ = where T is the torque at failure.
32800 10 10 509 N× ×
J
24
800 10 10 509 N mm20
32
τπ
= =⎛ ⎞×⎜ ⎟⎝ ⎠
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Problem 4: A solid brass rod AB (G=39 Gpa, 30 mm dia, 250 mm length) isb 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 TLJ L GJ
θ θ= ⇒ =A
BC
3
3
180 10 32027 10 164895.92Bθ
× ×=
× ×00.0129 rad 0.74= = Torque, T
180 Nm
4436 164895.92mm
32BCJ π ×= =
3
3
180 10 25039 10 79521 56ABθ × ×
=× ×
00.0145 rad 0.831= =
4430 79521.56mm
32ABJ π ×= =
39 10 79521.56× ×
0 000.74 0.83 . 711 1 5ACθ = + =
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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 theangles of twist between i) A and B; ii) A and C.
T G TLJ L GJ
θ θ= ⇒ =
3
3
400 10 120039 10 79521.56BAθ × ×
=× ×
0.1548 rad=
3
3
800 10 180039 10 251327.41CBθ − × ×
=× ×
0.1469 rad= −
4430 79521.56mm
32ABJ π ×= =θ θ θ= +
4440 251327.41 mmBCJ π ×
= =
32ABCA BA CBθ θ θ= +
0.1548 0.1469 0.0079 radCAθ = − =
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251327.41 mm32BCJ
Statically indeterminate shaftsy
A B
T T
BC
Torque, T
A BT T T= + (1)
A B
0Aθ = with respect to B
0θ θ− =
Torque, T
BC
0CB ACθ θ =
0B BC A CAT L T LG J G J
⇒ − =Torque, TReactive torque, TA
Reactive torque, TB
BC BC CA CAG J G J
(2)
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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
Stre
ss
Strain
• Corresponding shear stress distribution in the shaft:
τ
T rJ
τ =maxτ
R Entire cross-section is in the elastic range
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Materials in the inelastic ranges• Inelastic: Nonlinear stress strain curve
g
Stre
ss Elastic reboundon unloading
• Elastic plastic : Initially in the elastic range then fully plastic
Strain
• Elastic-plastic : Initially in the elastic range, then fully plastic
Stre
ss Elastic reboundon unloading
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S
Strain
Shear stress-strain curve Corresponding shear stress distribution in the shaft
ssmaxτ
Stre
s
Strain
R
ss
maxτ
Stre
s
Strain
R
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maxτ maxτFor any stress distribution, . .dT dA rτ=
T dT r dAτ= =∫ ∫R R
y ,maxτ
R
.A A
T dT r dAτ∫ ∫
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Problem 1:A solid steel shaft of 24 mm diameter is so severely twisted that only an 8 mmdiameter inner core remains elastic, while the rest of the diameter goes toinelastic 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 thesurface? Take G = 80 GPa.
τ
160 MPa4 mm
12 mm
0.002 γ φ=
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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 12160⎛ ⎞ ( )4 12
2 2
0 4
160 .2 . 160 .2 .4
r r dr r drπ π⎛ ⎞= +⎜ ⎟⎝ ⎠∫ ∫
( ) 3 316 558 10 Nmm 574 10 Nmm= + × = ×
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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 a tendency 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, but Outer 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.
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• 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 f applied torque. i.e., opposite to the direction of τmax .
• Similarly, all the inner regions at various radial distances have residual stresses stresses.
• These residual stresses are obtained as the difference between elastic-plastic stress distribution and the elastic stress distribution of rebound as shown in stress distribution and the elastic stress distribution of rebound, as shown in figure.
Elastic reboundτ ′E g : Elastic plastic material
Residual stressmaxτ
E.g.: Elastic-plastic material
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To find residual stresses on an elastic rebound, (i.e., on removal of the
T R′ 32574 10 12× ×
torque)
Shear stress on the surface T RJ
τ ′ = 24
574 10 12 211 N mm24
32π× ×
= =⎛ ⎞×⎜ ⎟⎝ ⎠
(considering elastic rebound)
Hence, residual shear stress on the surface2211 160 51 N mm= − =
maxτ τ′= −
160 MPa
Residual stress
211 MPa
12 mm4 mm
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0.006γ φ′ ′= =Residual twist on rebound
Strain variation is linear along the radius. 4 0.002φ =
Residual twist on rebound
θ12mmR =
4mmmax max
L RGθ τ
=At the initiation of yield,maxT T=
When T increases further to T’,
Lθ
increases beyond max , o tL L
θ θ ′ Strain variation
L L L
L rGθ τ=After yielding , is invalid from r=4 to 12.Note:
Final twist (after yielding,when T=T’)
θ ′
i.e., .L RGθ τ′ ′
≠ But, r
L r Rθ φ φ′ ′
= =
On elastic rebound, recovered twist Elastic Rebound(recovery of twist)
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TL GJθ ′=
Twist of a radial line
(recovery of twist)
To find residual twist on an elastic rebound,
3
0.0024 10
0.5 radm
m−= =×
r
L rθ φ′
=Final twist per unit length (after yielding,
when T=T’)L r
3 rad m1604
m 0.5 ra10
d8
m0
= =× ×
or, L rGθ τ′
=4 1080L rG
L rGθ τ=∵ is valid from r=0 to 4.
Elastic twist recovered (considering elastic rebound)
49 2 12 4
574Nm2480 10 N m 10 mπ −
=⎛ ⎞×
× × ×⎜ ⎟
( g )
TL GJθ ′= 0.22 rad m=
Hence, residual twist 0.5 0.22 0.28rad m= − =
80 10 N m 10 m32
× × ×⎜ ⎟⎝ ⎠
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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 of4.6 kNm. Assuming the shaft is made of an elastoplastic material with yieldstrength in shear of 150MPa and G = 80 Gpa, what are the radius of the elasticcore 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 andresidual (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
1504.6 10 N.mm .2 . 150 .2 .r r dr r drρ
ρ
π πρ
⎛ ⎞× = +⎜ ⎟
⎝ ⎠∫ ∫
( )3 336
2 2524.6 10 N.mm 150 1504 3
π ρπρ −× = +
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2815.743 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 =GL
τ θρ= (Valid from r = 0 to 15.743 only)
Strain variation along radius3
150 120080 10 15.743
LGτθρ
×∴ = =
× ×ρ
3142.92 10 rad−= ×
Final twist (after yielding,when T=T’)
θ ′
03 0180142.9 8. 82 0 11
π−= × × =
Elastic Rebound(reco er of t ist)
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29Twist of a radial line
(recovery of twist)
To find residual stresses on an elastic rebound, (i.e., on removal of the
T R′64.6 10 25× ×
torque)
Shear stress on the surface 2187 52 NT RJ
τ ′ = 45032
π=
⎛ ⎞×⎜ ⎟⎝ ⎠
(considering elastic rebound)2187.52 N mm=
Hence, residual shear stress on the surface
2187.52 150 37.52 N mm= − =
maxτ τ′= −
150 MPa
Residual stress
187.52 MPa
25 mm15.743 mm
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To find residual twist on an elastic rebound,
Elastic twist recovered (considering elastic rebound)
3
49 2 12 4
4.6 10 Nm 1.2m5080 10 N m 10 mπ −
× ×=
⎛ ⎞×× × ×⎜ ⎟
T LGJ
θ′
= 00.1124 rad 6.44= =
Hence residual twist 08 18 6 4 74 1 4= − =
80 10 N m 10 m32
× × ×⎜ ⎟⎝ ⎠
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)
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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 twistedthrough 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 oftorque?torque?
To find elastic core.
6 0.1047 rad180πθ = × = 145 MPa
GL
τ θρ= (Valid from r = 0 to ρ only)
16 mmρ10.792 mm=
ρ
3
145 600 10.792 mm77 10 0.1047
LGτρθ
×∴ = = =
× ×
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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 10.792145 1454 3
ππ −= +
1148475.884 NmmT ′∴ =
4 3
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To find residual stresses on an elastic rebound, (i.e., on removal of the
T R′′ 1148475.884 Nmm 16×
torque)
Shear stress on the surface2T R
Jτ ′ = 4
1148475.884 Nmm 1632
32π
=⎛ ⎞×⎜ ⎟⎝ ⎠
(considering elastic rebound) 2178.5N mm=
Hence, residual shear stress on the surface2178.5 145 33.5 N mm= − =
maxτ τ′= −
maxTRJ
τ= −Residual shear stress at 10.792 mmρ =
145 MPa
Residual stress
178.5 MPa
max4
1148475.884 Nmm 10.79232
τπ
×= −
⎛ ⎞×⎜ ⎟
16 mm10.792 mm2120.4 145 24.6 N mm= − = −
32⎜ ⎟⎝ ⎠
Dept. of CE, GCE Kannur Dr.RajeshKN
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 mminner diameter has been applied with a torque such that the inner surface firstreaches plastic zone, and then torque is removed. Steel is elastoplastic withyield 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 maximumresidual stress in the shaft and residual (permanent) angle of twist, afterremoval of torque?
30Applied 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 32 30 18145
3π −
=
145 MPa6428452.551 NmmT ′∴ =
( )4 460 36π × −18 mm 30 mm
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35
( ) 460 361107449.11 mm
32J
π ×= =
To find residual stresses on an elastic rebound, (i.e., on removal of the
T RJ
τ′
′ =6428452.551 Nmm 30
1107449 11×
=
torque)Shear stress on the surface(considering elastic rebound) J 1107449.11(considering elastic rebound)
2174.142 N mm=
Hence, residual shear stress on the outer surface2174.142 145 29.142 N mm= − =
maxτ τ′= −
τ ′
maxTRJ
τ= −Residual shear stress at 18 mmρ =
174 142 MPa=
145 MPa
Residual stress
τ
max6428452.551 Nmm 18
1107449.11τ×
= −
174.142 MPa
18 mm 30 mm 2104.49 145 40.51 N mm= − = −
Dept. of CE, GCE Kannur Dr.RajeshKN
Hence the maximum residual shear stress is on the inner surface.
To find residual twist on an elastic rebound,
Final twist per unit length (after yielding, when T=T’)
LrGτθ ′ =
L rGθ τ=∵ is valid from r=0 to 18.
3
145 1250 rad 0.131 rad18 77 10
×= =
× ×
Elastic twist recovered (considering elastic rebound)
8 77 0
3 2 4
6428452.551 Nmm 1250mm77 10 N mm 1107449 11mm
×=
× ×
Elastic twist recovered (considering elastic rebound)
T LGJ
θ′
= 0.094 rad=
Hence, residual twist 0.131 0.09 0.034 7rad= − =
77 10 N mm 1107449.11mm× ×GJ
00.037rad 2.1= =
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Bending moment and shear force
T f l d
g
Types of loadsPoint loadDistributed loadDistributed loadUniformly distributed loadUniformly varying loady y gCouple
T f tTypes of supportsFixed (built-in or encastre)Hinged (pinned)Hinged (pinned)RollerGuided fixed
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38Elastic (Spring)
Beams
Statically determinate and indeterminate beams
Beams
y
• Simply supportedSimply supported• Cantilever• Propped cantilever• Fixed• Continuous
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39
Sign conventions
Bending Momentg
S i H iSagging Hogging
Shear Force
Clockwise Anticlockwise
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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
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41
Bl
B
x
(+)P
SFD
(-)Pl−
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42BMD
lB
x
wl (+)SFDSFD
2wl− (-)
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43BMD
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
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44
C
P( ) 2
SFDC
A B
2P−
(+)
(-)SFD 2
4PL
A B
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45BMD C
2wL
(+)
2wL−
SF diagramCA
B2 (+)
(-)
SF diagram
2
8wL
A B(+)
2
8wL
A B(+)
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46BM diagramC
BM diagramC
Pba
x
Pb ( )L
SFDCA
BPaL
−
(+)
(-)SFD L
PabLA
B(+)
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BMD C
60 kN 50 kN
1
A D C B
1m3m
6m
x
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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
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49
P
bax
60 kN30 kN/m
5 kN/m25 kNm
A DCB E F
1m 2m1m 1.5m 2.5m
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50
Relation between shear force and bending momentRelation between shear force and bending moment
w
M M dM+A C
VV dV+
B D
0Y =∑0BM =∑
dx
0Y =∑
( ) . 0V V dV w dx− + − = ( ) ( )2
02
dxM w V dV dx M dM+ + + − + =
.dV w dx− =
dV−
0Vdx dM− =
dMV =
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51
wdx
= Vdx
=
Example 1Example 1
2xPxM =
2x
xdM P Vdx
= = Shear force
0xdVdx−
= Load intensity
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Example 2Example 2w kN/m
Lx L
2wL
2wL
2
2 2xwLx wxM = −
2
2x
xdM wL wx Vdx
= − = Shear force
xdV wdx−
= Load intensity
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dx
Example 3Example 3
lB
lx
x is taken in the negative direction2
2xwxM −
=
x is taken in the negative direction
Hence dx is negative2
x xx
dM dM wx Vdx dx
−= = = Shear forcedx dx−
x xdV dV wdx dx
−= = Load intensity
Shear force
Dept. of CE, GCE Kannur Dr.RajeshKN
54
dx dx−y
Shear force and bending moment diagrams by integrationShear force and bending moment diagrams by integration
dVw −= dMV =( ) 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= =
Dept. of CE, GCE Kannur Dr.RajeshKN
i.e.0 0, Vdx
Example 1
Total load = Wmaxw L W=
max
22
W
WwL
=
∴ =
L
L
3W 2
3W
dV− ( )∫2w W
To draw SFD
dVwdx
= ( ) 1V wdx C∴ = − +∫ max2
2At , w Wx w x xL L
= =
2
1 12 2
2 22
W W xV xdx C CL L
− −∴ = + = +∫
Dept. of CE, GCE Kannur Dr.RajeshKN
22
12
WxV CL
−∴ = +
W
W W
At 0, 3
Wx V= =
x
maxw W
1 103 3
W WC C∴ = + ⇒ =
max22
w x xL L
= =2
2 3Wx WVL
∴−
+= Parabolic variation
W3 2
3W−
SFD
Dept. of CE, GCE Kannur Dr.RajeshKN
3SFD
To draw BMD
dMVdx
=2
22 3Wx WM dx CL
⎛ ⎞−∴ = + +⎜ ⎟
⎝ ⎠∫⎝ ⎠
3
22i.e., 3 3Wx WxM CL
−= + +
3 3L
At 0 0x M= = 2i.e., 0 C=3Wx WxM −
∴ = +At 0, 0x M 2,23 3
ML
∴ = +
Cubic variation
i.e.0 0,dM V= =
For maximum bending moment,
29 3WL
i.e.0 0, Vdx
2
0Wx W L−+ ⇒
9 3
3L
Dept. of CE, GCE Kannur Dr.RajeshKN
BMD2 0
3 3x
L+ = ⇒ =
Example 2d
( ) 1V wdx C= − +∫PP
To draw SFD
∫
F t0 0LPortion ABL/4 L/4
A B CD
From to0 , 04
x x w= = =
V C∴ =L PPConstant1V C∴ =
But at 0, x V P= = 1C P⇒ = from 0 to, Lx xV P =∴ ==
Constant
, 1 from 0 to 4
, x xV P∴
3L L L
Portion BC
From to 3 , 04 4L Lx x w= = = Also, at , 0
4Lx V= =
3L L
Dept. of CE, GCE Kannur Dr.RajeshKN
3from to 4
0, 4L LxV x= =∴ = Constant
P
P−P−SFD
To draw BMD
Portion AB
2M Vdx C= +∫
2M Pdx C∴ = +∫ 2Px C= +From to 0 , 4Lx x V P= = =
Also, at 0, 0x M= = 2 0C⇒ = , from 04
to LM Px x x∴ = = =
Dept. of CE, GCE Kannur Dr.RajeshKN
Linear variation
Portion BC
2M Vdx C= +∫
Constant2M C∴ =From to 3 , 04 4L Lx x V= = =
, 4 4
Also, at L PLx M= = 2 4PLC⇒ = from 3,
4t
4o
4PL L LM x x∴ = = =
4 4 4
L L4
PL4
PL
BMD4 4
Dept. of CE, GCE Kannur Dr.RajeshKN
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 by bending moment diagrams shear force and bending moments by integration and by singularity functions
Dept. of CE, GCE Kannur Dr.RajeshKN
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