6. 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 use some basic formulas related to bearing stress and load distribution.
(a) Calculating the required diameter of the pins:
The bearing stress is given by the formula:
σ = P / (d * t)
where σ is the bearing stress, P is the load, d is the diameter of the pins, and t is the thickness of the plate.
In this case, the maximum allowable average bearing stress is given as 50 MPa, the maximum tensile load is 4.0 kN, and the thickness of the plate is 8.0 mm. We can rearrange the formula to solve for the diameter:
d = P / (σ * t)
Substituting the given values:
d = 4.0 kN / (50 MPa * 8.0 mm)
= (4.0 * 10^3 N) / (50 * 10^6 N/m^2 * 8.0 * 10^-3 m)
= 0.001 m = 1 mm
Therefore, the required diameter of the pins is 1 mm.
(b) Determining the stresses in the pins:
There are two types of stresses apparent in the pins: bearing stress and shear stress.
The bearing stress is the stress between the plate and the pins, calculated using the previously mentioned formula:
σ_bearing = P / (d * t)
Substituting the given values:
σ_bearing = 4.0 kN / (1 mm * 8.0 mm)
= (4.0 * 10^3 N) / (1 * 10^-3 m * 8.0 * 10^-3 m)
= 500 N/m^2 = 500 MPa
Therefore, the bearing stress in the pins is 500 MPa.
The shear stress is the stress taken by the pins due to the pulling force. The formula to calculate the shear stress is:
τ = P / (π/4 * d^2)
Substituting the given values:
τ = 4.0 kN / (π/4 * (1 mm)^2)
= (4.0 * 10^3 N) / (π/4 * (1 * 10^-3 m)^2)
= 1.27 * 10^9 N/m^2 = 1.27 GPa
Therefore, the shear stress in the pins is 1.27 GPa.