1. Mechanical Principles and Applications
Static Loading of Systems
Shear Stress and Strain

2. Today Aims Objectives To recall last weeks topics.
To highlight instances of shear loading.
To introduce the scientific concepts behind shear loading.
To introduce next weeks topic.
Objectives
To perform calculations relating to shear stress.
To perform calculations relating to double shear stress.

3. Re-cap of Previous Topic
Types of Loading
Tensile
Compressive
Shear
Torsion
Calculations
Stress Load/Area
Strain Change in Length/Original length

4. Shear Force Situations
Shearing Punching
Transverse Force Shear Pin

5. Shear Convention

6. Shear Strain Identification

7. Worked Example 1
Calculate the force needed to guillotine a piece of metal 3.2mm thick and 1.2m wide given that the ultimate shear stress is 50MPa.
Solution
L = 1.2m
t = m
 τ = 50 x 106 Pa
The area of the rectangle to be cut is m x 1.2m
A = x 1.2
A = m2
τ = F/A so F = τA = 50 x 106 x
= N
F = 192kN

8. Worked Example 2
Calculate the force needed to punch a hole 35mm diameter in a sheet of metal 2.5mm thick given that the ultimate shear stress is 60MPa.
Solution
The area to be sheared is the circumference x thickness
 d = 0.035m
t = m
τ = 60 x 106 Pa
A = πd x t = π x x
A = 275 x 10-6m2
τ =F/A so F= τA = 60 x 106 x 275 x 10-6
= 16493N
F = 16.5kN

9. Worked Example 3
Calculate the force needed to shear a pin 6mm diameter given that the ultimate shear stress is 60MPa.
Solution
The area to be sheared is the cross sectional area.
 R = 0.003m
τ = 60 x 106
A = πR2 = π x
A = 28.3 x 10-6 m2
F = τA = 60 x 106 x 28.3 x 10-6
= 1696N
F = 1.7kN

10. Self Assessment Ex 1. Solution 1
w = 0.7m
T = 0.004m .
τ = 45 x 106Pa
A = 0.7 x = m2
F = τ A
F = 45 x 106 x
F = N = 126kN

11. Self Assessment Ex 1. Solution 2
D = 0.025m .
T = m .
τ = 60 x 106Pa
A = πd x t = π x x = m2
F = τ A
F = 60 x 106 x
F = N = 7.539kN

12. Self Assessment Ex 1. Solution 3
R = 0.004m
τ = 55 x 106Pa
A = πR2 = π x = m2
F = τ A
F = 55 x 106 x
F = N = 2.764kN

13. Self Assessment Ex 1. Solution 4
F = 2.5 Tonnes
= 2.5 x 103 x 9.81
= 24525N
T = 0.003m
D =
τ = 75 x 106Pa
A = F/ τ = 24525/75 x 106 = m2
A = πdt so, d = A/πt = /(π x 0.003)
d = m
A 35mm hole cannot be punched.

14. Double Shear

15. Worked Example 4
A pin is used as a clevis in a supporting tie. The force in the tie will be a maximum of 70kN and the ultimate shear stress allowed in the pin is 100MPa. Calculate the minimum pin diameter necessary in order to achieve a Safety Factor of 4.
Solution
 Safety Factor = 4
τ = 100 x 106
The Working Stress = 100/4
τ= 25MPa
  F = 70 x 103N
The pin is in Double Shear so τ = F/2A
so A = F/2τ = 70 x 103/ (2 x 25 x 106)
A = m2
But A = πD2/4
So D = √(4A/π) = √(4 x /π)
= √
D = 0.043m (NOTE:- This has been rounded UP)

16. Self Assessment Ex 2. Solution 1
R = 0.005m .
τ = 70 x 106Pa
S.F. =
Working Stress (WS) = 70 x 106/2 = 35 x 106Pa
A = πR2 = π x = m2
WS = F/2A so F =2A x WS =2 x x 35 x 106
F = 5498N
F = 5.5kN

17. Self Assessment Ex 2. Solution 2
F = 3500N .
τ = 90 x 106Pa
A = F/2 τ = 3500/(2 x 90 x 106)
A = m2
A = πD2/4 so D2 = 4A/π = 4 x /π
D2 =
D = √
D = m
D = 5mm

18. Next Lesson Aims To recall this weeks topics.
To introduce the Modulus of Elasticity.
To perform Modulus of Elasticity calculations.
To introduce next weeks topic.