Chapter 6, Problem 6.2.8P

A plastic-lined steel pipe has the cross-sectional shape shown in the figure. The steel pipe has an outer diameter d₁= 100 mm and an inner diameter d₂= 94 mm. The plastic liner has an inner diameter d₁= 82 mm. The modulus of elasticity of the steel is 75 times the modulus of the plastic.

1. Determine the allowable bending moment M_allowif the allowable stress in the steel is 35 M Pa and in the plastic is 600 kPa.

If pipe and liner diameters remain unchanged, what new value of allowable stress for the steel pipe will result in the steel pipe and plastic liner reaching their allowable stress values under the same maximum moment (i.e., a balanced design)? What is the new maximum moment?

  

To determine

The allowable bending moment if the allowable stress for steel is 35 MPa and plastic 600 Kpa

Answer to Problem 6.2.8P

Allowable bending moment for plastic, M_{allowableplastic} = 1050.8 N-m

Allowable bending moment for Copper, M_{allowablesteel} =768.25 N-m

Explanation of Solution

Given:

Allowable stress for steel, s_steel = 35 MPa

Allowable stress for plastic, s_plastic= 600 kPa

d₁= 82 mm

d₂= 94 mm

d₃= 100 mm

E_steel= 75*E_plastic

Concept Used:

  $\begin{array}{l}{M}_{allowable,steel}=\frac{{\sigma }_{steel}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{steel}*\raisebox{1ex}{$d_3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ M_{allowable,steel}=\frac{{\sigma }_{steel}\pi \left(E_{steel}*\left(d_3{}^4-d_2{}^4\right)+E_{plastic}{d}_1{}^4\right)}{32*E_{steel}*d_3}\\ M_{allowable,plastic}=\frac{{\sigma }_{plastic}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{plastic}*\raisebox{1ex}{$d_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ M_{allowable,plastic}=\frac{{\sigma }_{plastic}\pi \left(E_{steel}*\left(d_3{}^4-d_2{}^4\right)+E_{plastic}{d}_1{}^4\right)}{32*E_{plastic}*d_2}\end{array}$

Calculation:

  $\begin{array}{l}{M}_{allowable,steel}=\frac{{\sigma }_{steel}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{steel}*\raisebox{1ex}{$d_3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ M_{allowable,steel}=\frac{{\sigma }_{steel}\pi \left(E_{steel}*\left(d_3{}^4-d_2{}^4\right)+E_{plastic}{d}_1{}^4\right)}{32*E_{steel}*d_3}\\ M_{allowable,steel}=\frac{35*{10}^6*\pi \left(\left({94}^4-{82}^4\right)+{100}^4\right)}{32*75*100}\\ M_{allowable,steel}=768.25N.m\end{array}$

  $\begin{array}{l}{M}_{allowable,plastic}=\frac{{\sigma }_{plastic}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{plastic}*\raisebox{1ex}{$d_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ M_{allowable,plastic}=\frac{{\sigma }_{plastic}\pi \left(E_{steel}*\left(d_3{}^4-d_2{}^4\right)+E_{plastic}{d}_1{}^4\right)}{32*E_{plastic}*d_2}\\ M_{allowable,plastic}=\frac{600*{10}^3\pi *75\left(\left({94}^4-{82}^4\right)+{100}^4\right)}{32*94}\\ M_{allowable,plastic}=1050.8N.m\end{array}$

Conclusion:

Allowable bending moment for plastic, M_{allowableplastic} = 1050.8 N-m

Allowable bending moment for Copper, M_{allowablesteel} =768.25 N-m

To determine

The value of the diameter of the copper rod for a balanced design

Answer to Problem 6.2.8P

The maximum moment is 1050.76 N-m in balanced condition

Explanation of Solution

Given:

Allowable stress for titanium, s_ti = 840 MPa

Allowable stress for titanium, s_cu= 700 MPa

Outer Diameter of the titanium rod, d₂ = 40 mm

E_ti= 110 GPa

E_cu= 120 GPa

Concept Used:

  $\begin{array}{l}Allowablebendingmomentofsteel=Allowablebendingmomentofplastic\\ \frac{{\sigma }_{steel}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{steel}*\raisebox{1ex}{$d_3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{{\sigma }_{plastic}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{plastic}*\raisebox{1ex}{$d_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Calculation:

  $Allowablebendingmomentofsteel=Allowablebendingmomentofplastic$

  $\frac{{\sigma }_{steel}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{steel}*\raisebox{1ex}{$d_3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{{\sigma }_{plastic}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{plastic}*\raisebox{1ex}{$d_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

  $\frac{{\sigma }_{Steel}{}_{allowable}}{E_{Stell}*d_3}=\frac{{\sigma }_{Plastic}{}_{allowable}}{E_{Plastic}*d_2}$

  ${\sigma }_{Steel}{}_{allowable}=\frac{600*{10}^3*100}{94}$

  ${\sigma }_{Steel}{}_{allowable}=47.87MPa$

  $Maximumbendingmoment=\frac{{\sigma }_{steel}\left(E_{steel}*I_{steel}+E_{plastic}{I}_{plastic}\right)}{E_{steel}*\raisebox{1ex}{$d_3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ $M_{\max }=\frac{47.87*{10}^6*75\left(\pi \left(\left({94}^4-{82}^4\right)+{100}^4\right)\right)}{*\raisebox{1ex}{$100$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ $M_{\max }=1050.76N.m$

Conclusion:

The maximum moment is 1050.76 N-m in balanced condition