The beam in the figures below has the cross section and loading indicated therein; determine for the most requested section, knowing that the allowable normal voltage is 1400 kgf/cm². a) Determine the sum of forces in X, Y  answer By=-1/4 and Ay=? For Ay I have two answers 1 or

Q: A у C n - Х 60° 40° B F = 800 lb

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Q: A simply supported wide-flange beam has the cross-section and loading as shown in the figure below.…

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Q: A у C n 60° х 40⁰ и B F = 800 lb

A: A force of 800 lb acting at 40° with rod which is inclined at angle 60° to the horizontal.

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A: T = 600 Nm GBC = 16 GPa or 16000 MPa GCD = 12 GPa or 12000 MPa dBC = 30 mm

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Q: A у C n 60° x 40° B F = 800 lb

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