Answer:
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
Answer:
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).
