### The Shear Stress–Strain Diagram

```Shear Stress and Strain
Shear Stress, Shear Strain, Shear Stress and Strain
Diagram
1
Shear Stress
 ave 
V
A
A: area of the plane
( the force is acted on the center of designated area)
Lecture 1
2
3
Shear Stress: pin and rivet
 ave 
VA
d
2
4
VA= 0.5 FA
4
Shear Stress: pin and rivet
What is the value of V and A?
5
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 =
6
Shear Stress: torque
V
T
V
Equilibrium equation:
T  0
T  VD  0
V 
T
D
7
Shear Stress: torque
Calculate the V and A.
Sketch the FBD.
T= 10 Nm ccw
Bolt d = 10mm
Position of the bolt R = 100 mm
8
Allowable Shear Stress
• The allowable stress is the maximum shear stress allowed due to the
properties of the material.
 all   ave
Considered as SAFE
 all   ave
Considered as FAIL
F .S 
 all
 ave
Safety Factor
Safe F.S >= 1
Fail F.S < 1
9
Shear Strain
Shear strain is detected the changes in angle is detected.
3 Normal strains (ex, ey, ez ) and 3 shear strains (gxy, gyz,
gxz) are detected.
10
Approximate length of the edges
(1  e x )  x
(1  e y )  y
(1  e z )  z
Approximate angles between three sides

2
 g xy

2
g

yz
2
 g xz
11
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.
  Gg
G = shear modulus of
elasticity
or the modulus of rigidity
• 3 material constants, E, and G are actually
related by the equation
G 
E
2 1  v 
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 curve
Yield shear stress: from curve
Ultimate shear stress: from curve
 
V
A
 V  A
Solution:
The coordinates of point A are (0.008 rad, 360 MPa).
Thus, shear modulus is
G 
360
0 . 008
  MPa
 45 10
3
(Ans)
By inspection, the graph ceases to be linear
at point A. Thus, the proportional limit is
 pl  360 MPa (Ans)
This value represents the maximum shear
stress, point B. Thus the ultimate stress is
 u  504 MPa (Ans)
Since the angle is small, the top of the will be displaced horizontally by
tan 0 . 008 rad   0 . 008 
d
 d  0 . 4 mm
50 mm
The shear force V needed to cause the displacement is
 avg 
V
A
;
360 MPa 
V
75 100 
 V  2700 kN (Ans)
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 u = 0.3.
Discuss the approach???
1) FBD of the bolt
V 
P
2
2) Ssy = 420 MPa, calculate the P
 y  420 
P/2
d / 4
2
P  263 . 9 N
3) G is slope of the curve
G  420 / 0 . 00545
 77 . 064 GPa
4) Calculate E based on value G and u
G 
E
2 1  v 
Review Question
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
Calculate the tension of cable AB, BD, and
BF
Calculate 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
20
Equilibrium
TBF
TBD
F
X
0
T BD cos( 29 . 5 )  T BF cos( 45 . 5 )
F
Y
0
T BD sin( 29 . 5 )  T BF sin( 45 . 5 )  9810  0
T BF  8840 N
W
T BD  7118 N
Stresses: the highest will be
cable BA as it has the highest
 all 
A
T
A
T
 all

9810
 70 . 1mm
2
140
d  9 . 44 mm
21
Internal forces at middle of BE (150mm)
TBF
F
Y
0
V  T BF  T BF sin( 45 . 5 )  0
P
V
V   2535 N
M
M
F
0
M  V * 150
TBF
 380250 Nmm
F
x
0
T BF cos( 45 . 5 )  P  0
P   6196 N
Lecture 1
22
Example
The joint is fastened together using two blots. Determine the required
Diameter of the bolts if the failure shear stress of the bolt is 350 Mpa.
Use factor of safety of F.S = 2.5.
23
V for each bolt
V  80 , 000 / 4  20 , 000 kN
Shear stress resulted from V
 
20 , 000
d / 4
2
Allowable shear stress
 all 


F .S
Therefore:
350
 140 MPa
2 .5
 all  
d  13 . 5 mm
 d  14 mm
24
Example
The A-36 steel wire AB has a
cross-sectional area of 10 mm2
and unstretched when q =45o.
needed to cause q = 44.9o.
25
Initial length LAB
L AB 
400  400 )
2
2
 565 . 68 mm
Final elongation
L ' AB 
400  400  2 ( 400 )( 400 ) cos( 90 . 2 )
2
2
 566 . 67 mm
Strain
e 
566 . 67  565 . 68
 1 . 75 (10 )
3
mm / mm
565 . 68
Normal Stress on cable
  E e  200 (10 ) (1 . 75 )(10 )
3
3
 350 MPa
26
The tension of the cable
 
T
A
T   A  350 (10 )  3500 N
FBD when the cable is stretched
O
M
O
0
 P cos( 0 . 2 )( 400 )  T sin( 44 . 9 )( 400 )  0
P  2 . 46 kN
P
27
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