A 35-ft long solid steel rod is subjected to a load of 8,000 lb. This load causes the rod to stretch 0.266 in. The modulus of elasticity of the steel is 30,000,000 psi. Determine the diameter of the rod (precision of 0.00)

Answers

Answer 1

Answer:

Here is the example how to do it

Explanation:

We can use the formula for the modulus of elasticity to solve for the diameter of the steel rod:

modulus of elasticity = (stress / strain) = (force / area) / (change in length / original length)

Solving for the area of the steel rod, we get:

area = (force * original length) / (modulus of elasticity * change in length)

Substituting the given values, we get:

area = (8,000 lb * 420 in) / (30,000,000 psi * 0.266 in) = 0.105 in^2

The area of a circle is given by the formula:

area = pi * (diameter/2)^2

Substituting the value we just calculated, we get:

0.105 in^2 = pi * (diameter/2)^2

Solving for the diameter, we get:

diameter/2 = sqrt(0.105 in^2 / pi) = 0.182 in

diameter = 2 * 0.182 in = 0.364 in

Therefore, the diameter of the steel rod is 0.36 in (to two decimal places).


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