3 years ago

The area under the moment diagram is shear force. a)-True b)-False

Engineering

1 answer:

liberstina [14]3 years ago

8 0

Answer:

False

Explanation:

as we know that $V(x)=\frac{dM}{dx}\\ \\=> M(x) = \int\limits^x_o {V(x)} \, dx \\\\$

=> Area under shear diagram gives the moment at any point but the reverse cannot be established from the same relation

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Answer:

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modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

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$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

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$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

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$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

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