Question

A hollow circular nylon pipe (see fig) supports a load PA = 1750 lb, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load PB is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d1 = 2 in., d2 = 2.375 in., d3 = 2.25 in., and d4 = 2.5 in., respectively. The upper pipe has a length L1= 14in., the lower pipe length is L2 = 16 in. Neglect the self-weight of the pipes. -If PA remains unchanged, find the new value of PB (in lb) so that upper and lower parts have same tensile stress.

Answer 1

Answer:

The value of $P_{B}$ is 4586.20 lb.

Explanation:

Given that,

Load = 1750 lb

The inner and outer diameters of the upper and lower parts of the pipe are

$d_{1}=2\ in$

$d_{2}=2.375\ in$

$d_{3}=2.25\ in$

$d_{4}=2.5\ in$

Length of upper pipe = 14 in

Length of lower pipe = 16 in

We need to calculate the value of $P_{B}$

Using tensile stress in upper segment

$\sigma_{1}=\dfrac{P_{A}+P_{B}}{A_{1}}$....(I)

Tensile stress in lower segment

$\sigma_{2}=\dfrac{P_{B}}{A_{2}}$.....(II)

From equation (I) and (II)

$\dfrca{P_{B}}{A_{2}}=\dfrac{P_{A}+P_{B}}{A_{1}}$

$P_{B}A_{1}=A_{2}P_{A}+A_{2}P_{B}$

$P_{B}(A_{1}-A_{2})=P_{A}A_{2}$

$P_{B}=\dfrac{P_{A}A_{2}}{A_{1}-A_{2}}$

$P_{B}=\dfrac{\dfrac{P_{A}}{A_{1}}}{\dfrac{1}{A_{2}}-\dfrac{1}{A_{1}}}$

$P_{B}=\dfrac{\dfrac{1750}{\dfrac{\pi}{4}(d_{2}^2-d_{1}^2)}}{\dfrac{1}{\dfrac{\pi}{4}\times(d_{4}^2-d_{3}^2)}-\dfrac{1}{\dfrac{\pi}{4}\times(d_{2}^2-d_{1}^2)}}$

Put the value into the formula

$P_{B}=\dfrac{\dfrac{1750}{\dfrac{\pi}{4}((2.375)^2-(2)^2)}}{\dfrac{1}{\dfrac{\pi}{4}\times((2.5)^2-(2.25)^2)}-\dfrac{1}{\dfrac{\pi}{4}((2.375)^2-(2)^2)}}$

$P_{B}=4586.20\ lb$

Hence, The value of $P_{B}$ is 4586.20 lb.