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## Homework Statement

Given the frame shown, determine the internal loadings at D

## Homework Equations

## The Attempt at a Solution

This should be a simple problem.....but I cannot see where I am making a mistake.

If I approach it by examining the entire structure, I get a wrong answer:

ΣFx: Ax + (cos60)IB = 0

**ΣFy**: Ay - (sin60)IB – 150 = 0

**ΣM about I**I: 3.44Ay – (7.44)(150) = 0

The 3.44 comes from determining, by trig, that IA is 3.44; the 150LB load is therefore 7.44 ft away from I

This yields Ay = 324.4

Plug back into sum of forces in y-direction; that yields IB = -201

OK now examine the section BDHC:

**ΣFx**: -(cos60)(201) + (cos45)GH=0

This yields GH= 142, which is wrong.

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However, if I start instead with just the structure BDHC, I get:

sum of moments about B: (2)[(sin45)GH] – (4)(150)=0 then GH=424 which agrees with the book

Where did I make a wrong turn? Any help greatly appreciated. Thank you.

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