Shear Stress and Strain

Shear Stress, Shear Strain, Shear Stress and Strain Diagram

1

Shear Stress

Lecture 1 2

A

Vave

V: shear load( the force is acted on the center of designated area)A: area of the plane

3

Shear Stress: Load (V)

4

Shear Stress: pin and rivet

42d

VAave

VA= 0.5 FA

5

Shear Stress: pin and rivet

What is the value of V and A?

6

Shear Stress: pin and rivet

What is the value of V and A?Sketch its FBD to show the value of V

V= 1 bolt: V= 2 parallel rivet V =3 parallel rivet V =

Vx= Vy =Total VR =

7

Shear load due to torsion

Shear Stress: torque

TVV

D

TV

VDT

T

0

0

Equilibrium equation:

8

Shear Stress: torque

T= 10 Nm ccwBolt d = 10mmPosition of the bolt R = 100 mm

Calculate the V and A.Sketch the FBD.

Allowable Shear Stress

• The allowable stress is the maximum shear stress allowed due to the properties of the material.

9

aveall Considered as SAFE

aveall Considered as FAIL

ave

allSF

.Safety FactorSafe F.S >= 1Fail F.S < 1

Shear Strain

Shear strain is detected the changes in angle is detected.

10

3 Normal strains (x, y, z ) and 3 shear strains (xy, yz, xz) are detected.

 

 

Approximate length of the edges

11

xx )1(

Approximate angles between three sides

 

 

yy )1( zz )1(

xy

2 yz

2 xz

2

The Shear Stress–Strain Diagram

• For pure shear, equilibrium requires equal shear stresses on each face of the element.

• When material is homogeneous and isotropic, shear stress will distort the element uniformly.

The Shear Stress–Strain Diagram

• For most engineering materials the elastic behaviour is linear, so Hooke’s Law for shear applies.

• 3 material constants, E, and G are actually related by the equation

G

G = shear modulus of elasticity or the modulus of rigidity

vE

G

12

Example

A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram is shown. Find the shear modulus G, the proportional limit, and the ultimate shear stress. Also, find the maximum distance d that the top of a block of this material could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement?

Example

Discuss the approach?

G: slope of the elastic curveYield shear stress: from curveUltimate shear stress: from curve

tan (0.008 rad) = d/50

AVA

V

Solution:

The coordinates of point A are (0.008 rad, 360 MPa).

Thus, shear modulus is

(Ans) MPa 1045008.0

360 3G

By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is

(Ans) MPa 360pl

This value represents the maximum shear stress, point B. Thus the ultimate stress is

(Ans) MPa 504u

Since the angle is small, the top of the will be displaced horizontally by

mm 4.0mm 50

008.0rad 008.0tan dd

The shear force V needed to cause the displacement is

(Ans) kN 270010075

MPa 360 ; VV

A

Vavg

Solve it!

The shear stress-strain diagram for a steel is shown in the figure. If a bolt is having a diameter of 20 mm is made of this material and used the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take = 0.3.

Discuss the approach???

2

PV

1) FBD of the bolt

2) Ssy = 420 MPa, calculate the P

NPd

Py

9.2634/

2/420

2

3) G is slope of the curve

GPaG 064.7700545.0/420

4) Calculate E based on value G and

vE

G

12

Review Question

20

A load W = 1000 kg is supported by a 3‐cable system as shown in figure 1. The cable run along 2 pulleys F and D which is attached to the solid wall by a round beam EF and CD both of which are 30mm diameter and 300mm long. Ignore the mass of cable and beam. All cables are the same size. The sizes of the pulleys at D & F can be ignored. Gravity = 9.81m/s2

Answer the following questions:Calculate the tension of cable AB, BD, and BFCalculate the diameter of the cable is the allowable stress is 140 MPa, assume all the cables have the same diameter.

Find the internal stress at the mid point of round beam FE

21

W

TBFTBD

Equilibrium

)5.45cos()5.29cos(

0

BFBD

X

TT

F

NT

NT

TT

F

BD

BF

BFBD

Y

7118

8840

09810)5.45sin()5.29sin(

0

Stresses: the highest will be cable BA as it has the highest load

mmd

mmT

A

A

T

all

all

44.9

1.70140

9810 2

Lecture 1 22

TBF

TBF

V M

P NV

TTV

F

BFBF

Y

2535

0)5.45sin(

0

Internal forces at middle of BE (150mm)

Nmm

VM

M F

380250

150*

0

NP

PT

F

BF

x

6196

0)5.45cos(

0

Example

23

The joint is fastened together using two blots. Determine the requiredDiameter of the bolts if the failure shear stress of the bolt is 350 Mpa. Use factor of safety of F.S = 2.5.

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V for each bolt

kNV 000,204/000,80

Shear stress resulted from V

mmd

mmdall

14

5.13

4/

000,202d

Allowable shear stress

MPaSFall 140

5.2

350

.

Therefore:

Example

25

The A-36 steel wire AB has a cross-sectional area of 10 mm2 and unstretched when =45o. Determine the applied load P needed to cause = 44.9o.

26

Final elongation

mm

L AB

67.566

)2.90cos()400)(400(2400400' 22

Initial length LAB

mm

LAB68.565

)400400 22

Strain mmmm /)10(75.168.565

68.56567.566 3

Normal Stress on cable

MPaE 350)10)(75.1()10(200 33

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The tension of the cable

FBD when the cable is stretched

O

P

kNP

TP

MO

46.2

0)400)(9.44sin()400)(2.0cos(

0

NATA

T

3500)10(350