Mechanics of structures module2

62
Mechanics of solids Mechanics of solids Torsion, Bending moment and shear force Bending moment and shear force Dr. Rajesh K. N. Assistant Professor in Civil Engineering Assistant Professor in Civil Engineering Govt. College of Engineering, Kannur 1

description

GCE Kannur

Transcript of Mechanics of structures module2

Page 1: 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 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

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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

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T

Torsion equation

R

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 →

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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

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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⎝ ⎠

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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

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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

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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

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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’,

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)

Page 27: Mechanics of structures  module2

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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Page 28: Mechanics of structures  module2

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 =∴

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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)

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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)

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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⎜ ⎟⎝ ⎠

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Hence the maximum residual shear stress is on the surface.

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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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( ) 460 361107449.11 mm

32J

π ×= =

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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= − = −

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Hence the maximum residual shear stress is on the inner surface.

Page 37: Mechanics of structures  module2

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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Page 38: Mechanics of structures  module2

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)

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Beams

Statically determinate and indeterminate beams

Beams

y

• Simply supportedSimply supported• Cantilever• Propped cantilever• Fixed• Continuous

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Sign conventions

Bending Momentg

S i H iSagging Hogging

Shear Force

Clockwise Anticlockwise

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Clockwise Anticlockwise

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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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Bl

B

x

(+)P

SFD

(-)Pl−

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42BMD

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lB

x

wl (+)SFDSFD

2wl− (-)

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43BMD

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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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C

P( ) 2

SFDC

A B

2P−

(+)

(-)SFD 2

4PL

A B

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45BMD C

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2wL

(+)

2wL−

SF diagramCA

B2 (+)

(-)

SF diagram

2

8wL

A B(+)

2

8wL

A B(+)

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46BM diagramC

BM diagramC

Page 47: Mechanics of structures  module2

Pba

x

Pb ( )L

SFDCA

BPaL

(+)

(-)SFD L

PabLA

B(+)

Dept. of CE, GCE Kannur Dr.RajeshKN

BMD C

Page 48: Mechanics of structures  module2

60 kN 50 kN

1

A D C B

1m3m

6m

x

Dept. of CE, GCE Kannur Dr.RajeshKN

48

Page 49: Mechanics of structures  module2

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

Dept. of CE, GCE Kannur Dr.RajeshKN

49

Page 50: Mechanics of structures  module2

P

bax

60 kN30 kN/m

5 kN/m25 kNm

A DCB E F

1m 2m1m 1.5m 2.5m

Dept. of CE, GCE Kannur Dr.RajeshKN

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Page 51: Mechanics of structures  module2

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 =

Dept. of CE, GCE Kannur Dr.RajeshKN

51

wdx

= Vdx

=

Page 52: Mechanics of structures  module2

Example 1Example 1

2xPxM =

2x

xdM P Vdx

= = Shear force

0xdVdx−

= Load intensity

Dept. of CE, GCE Kannur Dr.RajeshKN

Page 53: Mechanics of structures  module2

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

Dept. of CE, GCE Kannur Dr.RajeshKN

dx

Page 54: Mechanics of structures  module2

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

Page 55: Mechanics of structures  module2

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

Page 56: Mechanics of structures  module2

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

Page 57: Mechanics of structures  module2

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

Page 58: Mechanics of structures  module2

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+ = ⇒ =

Page 59: Mechanics of structures  module2

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

Page 60: Mechanics of structures  module2

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

Page 61: Mechanics of structures  module2

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

Page 62: Mechanics of structures  module2

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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