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11 April, 09:13

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.

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  1. 11 April, 10:08
    0
    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).
  2. 11 April, 12:13
    0
    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
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