A flat plate specimen of a soft metal alloy is to be tested by pulling on round pins inserted through the ends of the plate, as illustrated below
(a) Calculate the required diameter of the pins, given that the plate is 8.0 mm thick, the maximum tensile load is 4.0 kN and the maximum allowable average bearing stress is 50 MPa.
(b) What stresses are apparent in the pins?
To calculate the required diameter of the pins and determine the stresses in the pins, we need to consider the following information:
(a) Required diameter of the pins:
To calculate the required diameter of the pins, we can equate the maximum tensile load to the maximum allowable average bearing stress.
The average bearing stress can be calculated using the formula:
σ_bearing = F / (π * d * t)
Where:
σ_bearing is the average bearing stress
F is the maximum tensile load
d is the diameter of the pins
t is the thickness of the plate
Substituting the given values:
σ_bearing = 50 MPa = 50 x 10^6 N/m^2
F = 4.0 kN = 4.0 x 10^3 N
t = 8.0 mm = 8.0 x 10^-3 m
Rearranging the formula:
d = F / (σ_bearing * π * t)
Substituting the given values:
d = (4.0 x 10^3 N) / (50 x 10^6 N/m^2 * π * 8.0 x 10^-3 m)
Evaluating the expression:
d ≈ 3.183 mm
Therefore, the required diameter of the pins is approximately 3.183 mm.
(b) Stresses apparent in the pins:
The pins experience two types of stresses: bearing stress and shear stress.
- Bearing stress:
The bearing stress on the pins can be calculated using the formula mentioned earlier:
σ_bearing = F / (π * d * t)
Substituting the given values:
σ_bearing = (4.0 x 10^3 N) / (π * 3.183 x 10^-3 m * 8.0 x 10^-3 m)
Evaluating the expression:
σ_bearing ≈ 157.78 MPa
Therefore, the bearing stress in the pins is approximately 157.78 MPa.
- Shear stress:
The shear stress on the pins can be calculated using the formula:
τ = (F / A)
Where:
τ is the shear stress
A is the cross-sectional area of the pins, which is given by:
A = (π * d^2) / 4
Substituting the given values:
A = (π * (3.183 x 10^-3 m)^2) / 4
F = 4.0 x 10^3 N
Evaluating the expression for A:
A ≈ 8 x 10^-6 m^2
Substituting the values into the shear stress formula:
τ = (4.0 x 10^3 N) / (8 x 10^-6 m^2)
Evaluating the expression:
τ ≈ 5 x 10^8 Pa
Therefore, the shear stress in the pins is approximately 5 x 10^8 Pa.
In summary, the required diameter of the pins is approximately 3.183 mm, and the stresses apparent in the pins are approximately 157.78 MPa of bearing stress and 5 x 10^8 Pa of shear stress.