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## Homework Statement

The pin-connected structure is subjected to a load P as shown. Inclined member (1) has a cross-sectional area of 250mm

^{2}and a yield strength of 255MPa. It is connected to rigid member ABC with a 16-mm-diameter pin in a double shear connection at B. The ultimate shear strength of the pin material is 300MPa. For inclined member (1), the minimum factor of safety with respect to the yield strength is FS

_{min}=1.5. For the pin connections, the minimum factor of safety with respect to the ultimate strength is FS

_{min}=3.0.

Question: Based on the capacity of member (1) and pin B, determine the maximum allowable load P that may be applied to the structure.

**Ignore the angle written on the page, it's not the work that I used to solve the problem**

http://img593.imageshack.us/img593/6229/photobp.jpg [Broken]

## Homework Equations

[itex]\Sigma[/itex]F

_{x}=0

[itex]\Sigma[/itex]F

_{y}=0

[itex]\sigma[/itex]

_{allow}=[itex]\sigma[/itex]

_{failure}/(Safety Factor)

[itex]\tau[/itex]

_{allow}=[itex]\tau[/itex]

_{failure}/(Safety factor)

Safety Factor=[itex]\sigma[/itex]

_{failure}/([itex]\sigma[/itex]

_{actual})

Safety Factor=[itex]\tau[/itex]

_{failure}/([itex]\tau[/itex]

_{actual}

Safety Factor=P

_{failure}/(P

_{actual})

Safety Factor=[itex]\nu[/itex]

_{failure}/([itex]\nu[/itex]

_{actual})

## The Attempt at a Solution

http://img513.imageshack.us/img513/4791/photo1yi.jpg [Broken]

I know this is not the correct answer, and I know what the correct answer is

P

_{max}=15.03kN

Where did I go wrong?

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