The bars of the truss each have a cross-sectional area of 1.25 in2. if the maximum average normal stress in any bar is not to exceed 20 ksi, determine the maximum magnitude p of the loads that can be applied to the truss.

P=25000lbf

Explanation:

For this problem we will use the equation that relates, the effort, the area and the force for an element under normal stress.

σ=P/A

σ=stress=20kSI=20 000 lbf/in ^2

P=force

A=area

solving for P

P=Aσ

P=(20 000 lbf/in ^2)(1.25in^2)

P=25000lbf

The maximum magnitude P of the loads that can be applied to the truss = 25,000 Pounds or 111,205.5 Newtons.

Explanation:

In order to calculate the maximum load P, we will make use of the formula: Maximum average stress (20 ksi) = maximum load P ÷ cross-sectional area (1.25 in²)

Make P (the maximum load) the subject of the formula: P = 20 ksi × 1.25 in².

Before moving further, we have to convert the average normal stress (in ksi) to an appropriate unit: The average normal stress = 20 ksi = 20 kip per square inch (kip/in²)

But 1 kip = 1000 Pounds (i.e., 1000 lb)

Therefore, 20 ksi = 20,000 Pounds/in².

Therefore, P (maximum load) = 20,000 pounds/in² × 1.25 in² = 25,000 Pounds = 111,205.5 Newtons (because 1 Pound = 4.44822 Newtons).

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answer:

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explanation: