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Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for which the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B.

User Leang
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2 Answers

4 votes

Final answer:

The cross-sectional area required for link AB with an ultimate normal stress of 65 ksi and a factor of safety of 3.20 can be determined using the formula σ = P / A, where σ is the required stress of 20.3125 ksi. However, without knowing the actual load P, the exact cross-sectional area cannot be calculated.

Step-by-step explanation:

To determine the cross-sectional area of link AB for which the factor of safety will be 3.20 when the material's ultimate normal stress is 65 ksi, we use the following formula:

Required stress = Ultimate stress / Factor of safety

Thus, the required stress that the steel can withstand while ensuring a factor of safety of 3.20 is:

Required stress = 65 ksi / 3.20

Required stress = 20.3125 ksi

Now, to find the cross-sectional area (A), if the load (P) the link can carry at this stress is known, we use the stress (σ) formula:

σ = P / A

And rearrange it to solve for A:

A = P / σ

If P is unknown, additional information regarding the load the link needs to support is necessary to calculate the specific cross-sectional area. Without this information, one cannot provide a numerical value for A.

User Adam Sheehan
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Step-by-step explanation:

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User Vegashacker
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