In the figure, member (1) is a steel bar with a cross-sectional area of 1.3 in.2

Question

In the figure, member (1) is a steel bar with a cross-sectional area of 1.3 in.2 and a yield strength of 56 ksi. Member (2) is a pair of 6061-T6 aluminum bars having a combined cross-sectional area of 3.6 in.2 and a yield strength of 37 ksi. A factor of safety of 1.83 with respect to yield is required for both members. Determine the maximum allowable load P that may be applied to the structure. Report the factors of safety for both members at the allowable load. Answers: P = kips. FS1 = . FS2 = .

Explanation / Answer

Max force for aluminum = 37*3.6 = 133.2 kilopound

Max force for steel = 56*1.3 = 72.8 kilopound

Max horizon force for aluminum = 133.2 * cos31 = 114.2 kilopound

So the steel is the limiting (i.e. weaker) bar here. This means we give the steel a "safety factor" of 1.83 and

force on steel = 72.8 / 1.83 = 39.78

this will be the horizontal force on the aluminum, so

force on aluminum = 39.78/cos31 = 46.41 kilopound

so the safety factor for the aluminum will be 114.2/46.41 = 2.46

And the max allowable load will be

P = 46.41 * sin31 = 23.9 kips

So the three answers should be 23.9 kips..... FS1 = 1.83..... FS2 = 2.46

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