Question 1.5: A hollow circular nylon pipe (see Fig 1-23) supports a load ...
A hollow circular nylon pipe (see Fig 1-23) supports a load P_{A} = 7800 N, which is uniformly distributed around a cap plate at the top of the lower pipe. A second load P_{B} is applied at the bottom. The inner and outer diameters of the upper and lower parts of the pipe are d_{1} = 51 mm, d_{2} = 60 mm, d_{3} = 57 mm, and d_{4} = 63 mm, respectively. The upper pipe has a length L_{1} = 350 mm; the lower pipe length is L_{2} = 400 mm. Neglect the self-weight of the pipes.
(a) Find P_{B} so that the tensile stress in upper part is 14.5 MPa. What is the resulting stress in the lower part?
(b) If P_{A} remains unchanged, find the new value of P_{B} so that upper and lower parts have same tensile stress.
(c) Find the tensile strains in the upper and lower pipe segments for the loads in part (b) if the elongation of the upper pipe segment is known to be 3.56 mm and the downward displacement of the bottom of the pipe is 7.63 mm.
Numerical data:
- Inner diameter of upper pipe, d1 = 51 mm
- Outer diameter of upper pipe, d2 = 60 mm
- Inner diameter of lower pipe, d3 = 57 mm
- Outer diameter of lower pipe, d4 = 63 mm
- Load applied at the top, PA = 7800 N
- Length of upper pipe, L1 = 350 mm
- Length of lower pipe, L2 = 400 mm
Final Answer
(a) Find P_{B} so that the stress in the upper part is 14.5 MPa. What is the resulting stress in the lower part? Neglect self-weight in all calculations.
Use the given dimensions to compute the cross-sectional areas of the upper (segment 1) and lower (segment 2) pipes (we note that A_{1} is 1.39 times A_{2}). The stress in segment 1 is known to be 14.5 MPa.
The axial tensile force in the upper pipe is the sum of loads P_{A} and P_{B}.
Write an expression for \sigma_{1} in terms of both loads, then solve for P_{B}:
where \sigma_{1}=14.5 \mathrm{MPa} \text { so } P_{B}=\sigma_{1} A_{1}-P_{4}=3577 {N}
With P_{B} now known, the axial tensile stress in the lower segment can be computed as
\sigma_{2}=\frac{P_{B}}{A_{2}}=6.33 {MPa}(b) If P_{A} remains unchanged, find the new value of P_{B} so that upper and lower parts have same tensile stress.
So P_{A} = 7800 N. Write expressions for the normal stresses in the upper and lower segments, equate these expressions, and then solve for P_{B}.
Tensile normal stress in upper segment:
Tensile normal stress in lower segment:
\sigma_{2}=\frac{P_{B}}{A_{2}}Equate these expressions for stresses \sigma_{1} and \sigma_{2} and solve for the required P_{B}:
P_{B}=\frac{\frac{P_{A}}{A_{1}}}{\left(\frac{1}{A_{2}}-\frac{1}{A_{1}}\right)}=20,129 {N}So for the stresses to be equal in the upper and lower pipe segments, the new value of load P_{B} is 2.58 times the value of load P_{A} .
(c) Find the tensile strains in the upper and lower pipe segments for the loads in part (b).
The elongation of the upper pipe segment is \delta_{1} = 3.56 mm. So the ten-sile strain in the upper pipe segment is
The downward displacement of the bottom of the pipe is δ = 7.63 mm. So the net elongation of the lower pipe segment is \delta_{2}=\delta-\delta_{1}= 4.07 mm and the tensile strain in the lower pipe segment is
\varepsilon_{2}=\frac{\delta_{2}}{L_{2}}=1.017 \times 10^{-2}Note: As explained earlier, strain is a dimensionless quantity and no units are needed. For clarity, however, units are often given. In this example, ε could be written as 1017 × 10^{-6} m/m or 1017 μm/m.