3 years ago

A 5kg rock and a 10kg rock are dropped from a height of 10m?

1 answer:

5 0

Look out below ! You should step nimbly to one side, to avoid being hit by one or the other of those hazardous weight objects when they arrive (at the same time).

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A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$