Static shear and torsion test

1. STATIC SHEAR TEST

2. INTRODUCTION:
 The shear is the case of sliding one portion of the body
on the other portion at certain cross-section as a result
of shearing forces or twisting moments.
 The shear conditions that produced from shearing
forces are called “direct shear”
 While that produced from twisting moments are called
“torsional shear”.
 In all cases the shearing stress acts parallel to a plane,
as distinguished from tensile and compressive stresses
that acts normal to a plane.
THE DIRECT SHEAR
2

3. THE DIRECT SHEAR
3
An approximation of pure shear is the case of riveted joint. In this case
the upper half of the rivet is tending to slide over the lower half, and a
shearing stress is set up in the plane.
1- DIRECT SHEAR
The direct shear divided into:
A- Single Shear
(a) The shear stress is:
A
P


B- Double Shear
(b) The shear stress is:
A
2
P


C-Punching Shear
(c) The punching shear is :
r t
2π
P
τ 
Direct shear
a) Single Shear
b) Double Shear
c) Punching Shear

4. THE DIRECT SHEAR
4
2- SHEAR IN BEAMS
Ib
QS
τ 
 is the shearing stress
Q is the shearing force on the beam cross-section,
S is the first moment of area of the beam cross-section about the neutral
axis,
I is the moment of inertia of the beam cross-section, and b is the beam
width.
qmax

5. SHEAR STRAIN
 shear strain (γ) is an angular deformation measurement defined as
the change in a 90-degree angle at a point in the material during
loading
 Shear strain measurements are used in determining the shear
modulus G
 In the elastic range the shear stress ( ) is linearly
Proportional to shear strain (g) and the following
Relation governing the behavior:
 = G g
G: is the shear modulus or modulus of rigidity
which is defined as the resistance of material
to elastic shear deformation and related to the modulus of elasticity
by the following relation:
G = E/2(1+n)
Where: n is the Poisons ratio
THE DIRECT SHEAR
5

6. TORSION
3- TORSION SHEAR
The general torsion formula
L
Gθ
r
τ
J
T


Stress Due to Twisting
τ : shear stress N/m2
T: applied torque, N.m
J: Polar second moment of area, m4
θ: angle of twist, radian = θo ×( /180)
G: the shear modulus or modulus of rigidity, N/m2
r: the radius of the shaft, m
L: length of the shaft, m
6

7. TORSION TEST
7
ΘP.L
Te is the torque at elastic limit,
r is the radius of specimen, and
J is the polar second moment of area of the section
Torque-angle of twist diagram
TP.L
Mechanical properties in torsion
) Elastic shear strength, τe
Since
r
τ
J
T

Then,
J
r
e
T
e
τ 
J = d4/32 = r4/2
1- Elastic Shear Strength, e

8. TORSION TEST
2- Ultimate Shear Strength or modulus of rupture, max
For solid shaft
8
the torsional formula )
J
r
T
( max
max 
 gives maximum nominal stress larger
than the true maximum stress. Therefore the modulus of rupture can be
calculated from the following empirical equations:
For ductile material 3
πd
max
12T
max
τ 
For brittle material 3
πd
max
14T
max
τ 
t
r
2π
max
T
max
τ 2

Where: t is the wall thickness, and
r is the mean radius
For hollow shaft

9. 3- Shear Modulus or Modulus of rigidity, G
TORSION TEST
L
Gθ
J
T

Jθ
TL
G 
Where: T is the torque in the elastic range,
 is the angle of twist (in radian) in the elastic range, and
L is the length of specimen.
Modulus of rigidity, G, is similar to Young's modulus of elasticity, E
For most materials E is about 2.5 times greater than G.
υ)
2(1
E
G


Where,  is the Poisson's ratio
9

10. TORSION TEST
10
d) Ductility
The ductility is expressed by the percentage elongation of the specimen outer
fiber.
Where L
is the final fiber length and is computed knowing L and r and L is
the original gage length.
x100
L
L
L
e
%


4- Ductility
Stress Due to Twisting
L
L
L= √ L2+(rθ)2

11. TORSION TEST
5- Resilience
Resilience = (½) TPLPL
Modulus of resilience = Resilience / volume
6- Toughness
Toughness = (2/3) Tmaxmax (at rupture)
Modulus of Toughness = Toughness / Volume
TP.L
Tmax
θmax 11

12. TORSION TEST
12
7- Failure under torsion
Ductile materials, such as steels, which break in shear in the torsion test, the
break in solid bars is plane normal to the axis of the specimen
Brittle materials, such as cast iron, fail on planes of maximum tension, which
occur at 45o to both the specimen axis and the specimen surface. This produces
a helical spiral fracture
𝐒𝐡𝐞𝐚𝐫 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡
𝐓𝐞𝐧𝐬𝐢𝐥𝐞 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡
= 0.8 (for ductile)
= 1.1- 1.3(for brittle)

13. TORSION
13
Non Uniform and Composite Shafts
For continuous shaft having two different diameter
T = T1 = T2
 = 1 + 2
For concentric shaft
T = T1 + T2
 = 1 = 2
Power = 2 N T
P(Watt),N( rev./sec)and T(N.m)
Power = 2 N T /75
P(hp),N( rev./sec)and T(kg.m)

14. Example 1
Rivets
Rivets
of
No
cm
ton
d
A
P
4
55
.
3
84
.
0
98
.
2
.
/
98
.
2
)
6
.
1
(
4
6
4
/
6 2
2
2











Solution
A- Single Shear
B- Double Shear
Rivets
Rivets
of
No
cm
ton
d
A
P
2
77
.
1
84
.
0
49
.
1
.
/
49
.
1
)
6
.
1
(
2
6
)
4
/
(
2
6
2
2
2
2











14

15. Example 2
2
4
. /
8
.
2486
32
/
)
6
.
1
(
8
.
0
2000
cm
kg
J
Tr
l
p 





2
2
4
.
.
/
48
.
142
/
9
.
142482
20
)
6
.
1
(
180
32
16
2000
cm
t
cm
kg
J
L
T
G
l
p
l
p












2
2
max
max
/
56
.
976
16
)
6
.
1
(
180
3
4
600
4500
2
3
/
2
. cm
kg
L
A
T
T
M 















2
3
3
max
max /
47
.
4196
)
6
.
1
(
4500
12
12
cm
kg
d
T







Solution
15

16. 2
4
. /
26
.
377
32
/
)
3
(
5
.
1
2000
cm
kg
J
Tr
l
p 





2
2
4
.
.
/
4
.
14
/
12
.
14410
20
)
3
(
180
32
20
2000
cm
t
cm
kg
J
L
T
G
l
p
l
p












2
.
. /
63
.
188
2
26
.
377
.
cm
kg
S
F
l
p
stress
oper 




2
3
3
max
max /
6
.
636
)
3
(
4500
12
12
cm
kg
d
T







16

17. The break in the solid bar is plane normal to the axis of the specimen
17

18. THANK YOU