Concept explainers
In the structure shown, an 8-mm-diameter pin is used at A, and 12-mm-diameter pins are used at B and D. Knowing that the ultimate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired.
Fig. P1.55
The allowable load P when an overall factor of safety of 3.0 is desired.
Answer to Problem 55P
The allowable load P when an overall factor of safety of 3.0 is desired is
Explanation of Solution
Given information:
The diameter (d) of each pin B and D is
The diameter (d) of pin A is
The ultimate shearing stress
The ultimate normal stress
Calculation:
Sketch the free body diagram of ABC as shown in Figure 1.
Refer to figure 1.
Take a moment about B.
Take a moment about A.
Find the area of double shear pin at A using the relation:
Substitute
Find the value of
Substitute
Find the value of P using the relation:
Substitute
Find the area of double shear pin at B and D using the relation:
Substitute
Find the force in member BD based on double shear in pins at B and D using the relation:
Substitute
Find the value of P using the relation:
Substitute
Find the area based on compression in links BD for one link as follows:
Here, d is the diameter of pin and b is the width of the section.
Substitute
Find the force in member BD of pin at B and D for one link using the relation:
Here, A is the area based on compression in link BD.
Substitute
Find the value of P using the relation:
Substitute
Based on results,
Select the smaller value of P is
Thus, the allowable load P when an overall factor of safety of 3.0 is desired is
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Chapter 1 Solutions
Loose Leaf For Mechanics Of Materials Format: Looseleaf
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