Quick review

Statically Indeterminate (Sections 4.4 & 4.5):

Statically Indeterminate, Thermal Strain, Poisson’s Ratio

number of unknowns > number of the equilibrium equations

Additional equation is needed to solve the problem

Thermal Strain (Section 4.6):

TLT TLT

T

Poisson’s ratio (Section 3.6):

allongitudin

lateral

)1(2

EGg

Lecture 3

Chapter 7: Torsion (Sections 5.1~5.4)Internal torque in a shaft

Relating the stress in a shaft to its angle of twist

Relating to the shear stress to the internal torque in the shaft

A sign convention for shaft torsion

Torque diagrams

Ch t 8 BM & SF di b ilib i (S tiChapter 8: BM & SF diagrams by equilibrium (Section

6.1)Point Forces and Couples

Distributed Forces

Chapter 7 TorsionFor the drive shaft in a rear wheel driveR For the drive-shaft in a rear wheel drive car, the gearbox applies a twisting moment or torque to one end of the h ft d th diff ti l li

Rear wheel

Engine

w

Shaft Differentialshaft, and the differential applies an opposite torque at the other end.

‘Cut’

Internal torque in a shaft

veM

0

TTTT intint 0

Relating the stress in a shaft to its angle of twist

L

Relationshipβ

LB

ρ is radius of the shaft;L is the length of the shaft

A

ρ gΦ is the angle of twist under the internal torque T.

A’ Ф

O

Shear strain on the surface of the shaft is given by

L

surface

Relating the stress in a shaft to its angle of twistr

r

surfacer radiusat

r

L

surface

Lr r radiusat

Untwisted Twisted GHooke’s Law

LGr

Note: Shear stress and shear strain are greatest on the surface of the shaft and

L

Shear stress and shear strain are greatest on the surface of the shaft and zero at the centre.

Relating the shear stress to the internal torque in a shaftdAdF

drrddA G L

Gr

drrdL

GrdF

dFdM

ddGrdM 3rdFdM ELEMENT

drdL

dM ELEMENT

2

ELEMENTHOOP dMdM

drL

GrdM 3

HOOP 2

0

ELEMENTHOOP

L

4

0HOOPdMM

2

4J

GL

M2

4

L

JGM

2

Relating the shear stress to the internal torque in a shaft

MT JGMT

LJGT

Torque is a moment that twists a member about its longitudinal axis.

Where J is called the polar moment of area of the shaft It is a geometric property of the cross-section

g

shaft. It is a geometric property of the cross section of the shaft.

Solid shaft

2

4J ρ is the radius of the shaft2

Hollow shaft

ρ is the radius of the shaft.

ρi

ρo

Hollow shaft

)( 44 2

)( ioJ

ρo is the outer radius of the shaft.ρi is the inner radius of the shaft.

SummaryRelationship between the shear stress τ and the angle of twist Φ

Gr L

JGRelationship between the internal torque T and the angle of twist Φ

LJGT

The Engineers’ Theory of Torsion is as follows

TLL

L is the length of the shaftG is the shear modulusJ is the polar moment of area

GJGr J is the polar moment of area

r is the radius at which we want to calculate the shear stress

A sign convention for shaft torsion

When you look at the shaftWhen you look at the shaft from an end, the end closer to you is twisting positively.

We cut the shaft to investigate the internal torque in a shaft with a positive twist.

We rotate our view of each portion, so that we are looking straight onto the cut surface. We see that a positive internal torque appearssee that, a positive internal torque appears anticlockwise.

When the shaft is viewed from the side, Tintand Φ are positive if, the torque at the left hand

A B

T T p , qend appears to act upward, and the torque at the right end appears to act downward.

T T

For multiple torques, To determine the magnitude and sign of the internal torque at Point P, between B and C.q

Using equilibrium to determine Tint.

S ti A B P

0M

Section A-B-P:

ve

NmTT 501015 intint

Section P-C-D:

ve 0

M

NmTT 505 intint

For multiple torques,Using the similar procedure, we can determine the internal torque at other portions of the shaftthe internal torque at other portions of the shaft.

NmT 10int Between A and B:

Between B and C: NmT 5int

Between C and D: NmT 0int

downward"point"thatPofleftthetotorquesAll

upward"point " that P, ofleft the to torquesAllintT

downwardpoint that P,ofleft theto torquesAll

Point P between B and C:

NmT 51510int

Torque Diagram,NmT 10int Between A and B:

Between B and C: NmT 5int

Between C and D: NmT 0int

Draw a diagram of the internal torque vs distance (x) along the shaft,

A

Tint

Draw a diagram of the internal torque vs distance (x) along the shaft,

We move from left to right, anA B

C D

We move from left to right, an “upwards” torque causes an increase in the torque diagram and a “downwards” torque causes ax a downwards torque causes a decrease in the torque diagram.

Torque Diagram,Two points A and B are separated by a distance x. The internal torque (T), shear modulus (G), and polar moment of area (J) are all constant between A and B. The angle of twist of B relative

Torque diagram

gto A is as follows

Tx

A

Tint

Torque diagramJGAB /

TTAB T

ABJGTxx

JGT

/Area

x

JGT is equal to the area under a graph of vs x

between A and B.JGTx

JGT

JGT

x B andA between x vsJGT ofgraph aunder Area/ ABx JG

Weighted torque diagram

Example 1, To determine the angles of twist of B relative to A, C relative to A

0.5mL ,70 ,9m-1EJB,andA Between

4 GPaG

Torque diagram0.6mL ,60 ,9m-2EJ

C, and BBetween 4 GPaG

Torque diagram T

d07140501430

10 Between A and B,

radAB 0714.05.0143.0/ -5 x T/JG

B t A d C0.143

AC 6.00417.05.0143.0/

Between A and C,

0.5m

-0.0417 x

0.6m rad0464.0

Example 2, The torques shown are exerted on pulleys B and C. The shafts are solid,• AB is made of aluminium (G=28GPa) with a diameter of 30mm, and• BC is made of steel (G=77GPa) with a diameter of 40mm.Determine the angle of twist of C relative to B, and C relative to A. Also determine the a d C e a e o so de e e emaximum magnitude of the shear stress, and where it occurs.

Torque diagram—T vs x

Weighted Torque diagram—T/JG vs x

TxArea under the weighted Torque diagram

JGTx

AB /

Step 1, Determine the support reaction at the wall

A C

Twall

B

400Nm 300Nm

Use equilibriumUse equilibrium,

ve 0

M

NmTT wallwall 1000300400

Step 2, Construct a torque diagram for this problem

TUse “cut’ and equilibrium300Nm

A

C Use cut and equilibrium,

NmTBC 300-100Nm xB

Step 3, Determine the values of shear modulus (G) and polar moment of area (J) for each part of the shaft

444 )2/30(142.3

Between A and B,

48953.72

)2/30(142.32

mEJ

2)3227.2()928()8953.7( NmEEEJG

444

)75142()2/40(142.3 mEJ

Between B and C,

)7514.2(2

)(2

mEJ

2)4935.1()977()7514.2( NmEEEJG

Step 4, Weight the torque diagram by 1/JG

T/JG

0.0155

1.2m 0.9m

-0.0449 x

Step 5, the rotation of C relative to B is equal to the area under the weighted torque diagram between B and C.

0.0155

T/JG

1.2m

-0.0449 x

1.2m

0.9m

radBC 0140.09.00155.0/

Step 6, the rotation of C relative to A is equal to the area under the weighted torque diagram between A and C.

radAC 0399.02.10449.09.00155.0/

Th h i l di ti f th t ti i b th i f Φ dThe physical direction of the rotation are given by the signs of ΦC/B and ΦC/A, and the sign convention described earlier.

Step 7, The maximum shear stress can be calculated by rearranging the Engineers’ Theory of Torsion

GJTL

GrL

L

GrJTr

GJGr

O th t id f h ft ti AB

LJ

MPaEE

JT 861.18

)89537(2/)330(100

On the outside of shaft portion AB:

EJ )8953.7(

On the outside of shaft portion BC:

MPaEE

JT 866.23

)7514.2(2/)340(300

Power: shafts are often to transmit power.

TP TP ω is the angular velocity of the shaft in radians/second.

Chapter 8 Bending Moment and Shear Force DiagramsTo determine stresses in a beam being bent is to find the internal bendingTo determine stresses in a beam being bent is to find the internal bending moment and shear force being carried.The procedures for construction bending moment and shear force diagrams were presented in the unit Fundamentals of Mechanicswere presented in the unit Fundamentals of Mechanics.

Example 1—Point forces and couples 10kN 4kNm

3.5m2m

4kNmDraw shear force and bending moment diagrams for the beam shown.

1. Cutting the beam2 Draw the free body diagram

5m

2. Draw the free body diagram3. Use equilibrium

0M0F0 F

4. Find internal bending moment and h f

0M ,0F ,0 Ay xF

shear force

Step 1: Calculate the support reactions 10kN 4kNm

Use equilibrium.

3.5m5m

2m

Use equilibrium.

0H0

vexF

10kN 4kNm

H

0100

BAvey RRF

3.5m5m

2m RB RA 0421050

Bve

A RM

4210 kNRB 8.45

4210

kNR 258410 kNRA 2.58.410

Step 2: Identify every point where the loading on the beam changes

At x=2m, a point force 10kN is applied

At x=3 5m a point couple 4kNm is applied

mxmmx5.32

20

At x 3.5m, a point couple 4kNm is applied mxm 55.3

Step 3: Make an imaginary cut (0<x<2m) 10kN 4kNm

H

x

Make an imaginary cut through the beam at K. The internal forces include 3.5m

5m

2m RB RA K

a normal (axial) force Nx

a shear force vxK

Mx

a shear force vx

a bending moment Mx x vx

Nx

5.2

The subscripts x indicate that Nx, vx and Mxmay be functions of the distance along the beambeam.

Sign Convention for Internal ForcesForces

Sign convention is established for internal forces for Sign convention is established for internal forces forthe consistency in analysisInternal forces at a specified pointInternal forces at a specified point

Bending Moment

Axial Force

Shear Force

Axial force

(+) Axial force

Sign Convention for Internal Forces

Shear force

(+) Shear force

Bending moment

(+) Bending moment( ) g

Use the equilibrium equations to find expressions for Nx, vx and Mx

10kN 4kNm

H

x

3.5m5m

2m RB RA K00

xvex NF

0250 F

KMx

02.50

xvey vF

kNvx 2.5

02.50

xMM xve

Kx

Vx

Nx

5.2

kNM 25

Step 4: Make an imaginary cut (2m<x<3.5m)

xkNmM x 2.5

2

10kN 4kNm

RR

H

x

K3.5m

5m

2m RB RA K0102.50

xvey vF

kK

Nx

5 2

Mx 10

0)2(102.50 xxMM xK

kNvx 8.4

xVx

5.2ve

kNmxM x )208.4(

Step 5: Make an imaginary cut (3.5m<x<4m) 10kN 4kNm

H

x

3.5m5m

2m RB RA

H

K0102.50

xvey vF

kNv 8.4 5m

KMx10 04)2(102.50

xxMM x

veK

kNvx 8.4

x Vx

Nx

5.2 4

kNmxM x )248.4(

Step 6: Constructs graphs of shear force and bending momentShear force Bending moment

vx=5.2 vx Mx=5.2x Mx Mx=-4.8x+20

Mx=-4.8x+24

x vx=-4.8 x

Distributed Forces wx

The intensity of the distributed force can beThe intensity of the distributed force can be constant or can vary along the beam, and is given the symbol ωx.Note: ωx Is defined to be positive if it is acting downward, and it has units of force per length (N/m)force per length (N/m). The subscript x indicates that the intensity can be a function of distance along the beam (x). The equivalent point force of the distributed force Feq has a magnitude equal to the area of the region under a graph of ωx versus x. The line of action of Feq passes through the centroid of this region.action of Feq passes through the centroid of this region.

Feq=wL

Uniformly distributed force (ωx=constant)Feq=wmaxL/2

Linearly varying force

w

Feq wLwmax

L/2L L

L/3

Example 2: Distributed Force3kN/m

Find expressions for the shear force and2m

x

A B

C

Find expressions for the shear force and bending moment being carried by the beam shown, as functions of the distance from the l ft h d d ( )

5mx Cleft hand end (x).

1. Cutting the beam2 Draw the free body diagram2. Draw the free body diagram3. Use equilibrium

0M0F0 F

4. Find internal bending moment and h f

0M ,0F ,0 Ay xF

shear force

Step 1: Replace the distributed force by an equivalent point force

3kN/ 10kN Feq=9kN

2m

3kN/m

A B

10kN Feq 9kN

H

5m x C

5m 1.5m RBRA

Step 2: Use the equivalent point force, calculate the support reactions

0H0

vexF

090

BAvey RRF

053950 RM 05.3950

Bve

A RM

kNRB 3.65.39

B 5

kNRA 7.23.69

2m

3kN/m

Step 3: Identify every point where the loading on the beam changes

5mx

A B

CK

loading on the beam changes

The loading changes where the distributed force starts at point C (at

KMx

p (x=2m).

Step4: Make an imaginary cut (0<x<2m)x

Vx

Nx

2.7

Step4: Make an imaginary cut (0 x 2m)

Use the equilibrium equations to find expressions for Nx, vx and Mx

00

xvex NF

07.20

xvey vF

kNvx 7.2

07.20

xMM xve

K

x

xkNmM x 7.2

2m

3kN/m

Step 4: Make an imaginary cut (2m<x<5m)

5mx

A B

C

Step 4: Make an imaginary cut (2m x 5m)

0)2(37.20

xvey vxF K

0)2(32720

xxxMM

kNxvx )37.8(

2

3kN/m

0)2(32

7.20

xxMM xve

K

kNmxxM x )67.85.1( 2 vx

K

2m

x

A

C2 7

Mx

5m2.7

2m

Feq=3(x-2)

vx

K

2m

A

2.7

Mx

(x-2)/2

2.7

Step 5: Constructs graphs of shear force and bending momentSh f Bending momentShear force

vx=2.7

vx

Bending moment

Mx=2.7x Mx Mx=-1.5x2+8.7x-6

x

x xvx=8.7-3x

Note

1. Make a cut and a new free body diagram for every change in the beam’s loading.

2 Unfortunately this can make drawing shear force and bending moment2. Unfortunately, this can make drawing shear force and bending moment diagrams very tedious. Chapter 9 and 10 will look at shortcut ways to draw these.

f3. When determining the shear force and bending moment under a distributed force, always cut the beam and replace the distributed force by an equivalent point force.

4. We could use a free body diagram of section of the beam on the right hand side of the cut point (K) instead of the left hand side, as used aboveabove.a. Since every action has an equal and opposite reaction, the positive

directions for vx and Mx on the right hand portion are equal and opposite to those on the left hand portion.

b. The length of the portion is equal to the overall length minus x.