Category Archives: Strength of Materials

Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy A. Philpot Problem P1.2

A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the axial normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution:

We are given the following values:

Outside Diameter,D=2.50 inAxial Load,P=27 kipsMaximum Axial Stress,σ=18 ksiInside Diameter,d=D−2t\begin{align*} \text{Outside Diameter}, D & = 2.50\ \text{in} \\ \text{Axial Load}, P & = 27\ \text{kips} \\ \text{Maximum Axial Stress}, \sigma & = 18\ \text{ksi}\\ \text{Inside Diameter}, d & = D-2t \end{align*}

We are solving for the unknown wall thickness of the tube, tt.

From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi

σ=PAAmin=PσAmin=27 kips18 ksiAmin=1.500 in2\begin{align*} \sigma & =\frac{P}{A} \\ A_{\text{min}} & =\frac{P}{\sigma } \\ A_{\text{min}} & = \frac{27\:\text{kips}}{18\:\text{ksi}} \\ A_{\text{min}} & = 1.500\:\text{in}^2 \end{align*}

The cross-sectional area of the aluminum tube is given by

A=π4(D2−d2)A=\frac{\pi }{4}\left(D^2-d^2\right)

Set this expression equal to the minimum area and solve for the maximum inside diameter, dd

A=π4(D2−d2)A=π4[(2.50 in)2−d2]1.500 in2=π4[(2.50 in)2−d2]4(1.500 in2)=π[(2.50 in)2−d2]4(1.500 in2)π=(2.50 in)2−d2d2=(2.50 in2)−4(1.500 in2)πd=(2.50 in2)−4(1.500 in2)πd=2.08330 in\begin{align*} A & =\frac{\pi }{4}\left(D^2-d^2\right) \\ A & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 1.500\ \text{in}^2 & = \frac{\pi }{4}\left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ 4 \left( 1.500\ \text{in}^2 \right) & = \pi \left[\left(2.50\ \text{in}\right)^2-d^2\right] \\ \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} & = \left(2.50\ \text{in}\right)^2-d^2 \\ d^{2} & = \left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi} \\ d & = \sqrt{\left( 2.50\ \text{in}^{2} \right) - \frac{4 \left( 1.500\ \text{in}^2 \right)}{\pi}} \\ d & = 2.08330\ \text{in} \end{align*}

The outside diameter DD, the inside diameter dd, and the wall thickness tt are related by

D=d+2tD = d+2t

Therefore, the minimum wall thickness required for the aluminum tube is

t=D−d2t=2.50 in−2.08330 in2t=0.208 in  (Answer)\begin{align*} t & =\frac{D-d}{2} \\ t & = \frac{2.50\:\text{in}-2.08330\:\text{in}}{2} \\ t & = 0.208\ \text{in} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Mechanics of Materials: An Integrated Learning Approach 3rd Edition by Timothy Philpot Problem P1.1

A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the axial normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution:

We are given the following values:

Outside Diameter,D=60 mmWall Thickness,t=5 mmInside Diameter,d=D−2t=60 mm−2(5 mm)=50 mmMaximum Axial Stress,σ=200 MPa=200 Nmm2\begin{align*} \text{Outside Diameter}, D &= 60\ \text{mm} \\ \text{Wall Thickness}, t & = 5\ \text{mm} \\ \text{Inside Diameter}, d & = D-2t = 60\ \text{mm}-2\left( 5\ \text{mm} \right) = 50\ \text{mm}\\ \text{Maximum Axial Stress}, \sigma & =200\ \text{MPa} = 200\ \frac{\text{N}}{\text{mm}^2} \end{align*}

The cross-sectional area of the stainless-steel tube is

A=π4(D2−d2)A=π4[(60 mm)2−(50 mm)2]A=863.938 mm2\begin{align*} A & = \frac{\pi}{4}\left( D^2 - d^2 \right) \\ A & = \frac{\pi}{4}\left[ \left( 60\ \text{mm} \right)^2 - \left( 50\ \text{mm} \right)^2 \right] \\ A & = 863.938\ \text{mm}^2 \end{align*}

The normal stress in the tube can be expressed as

σ=PA\sigma =\frac{P}{A}

The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P.

Pmax=σmaxAPmax=(200 MPa)(863.938 mm2)Pmax=(200 Nmm2)(863.938 mm2)Pmax=172788 NPmax=172.8 kN  (Answer)\begin{align*} P_{max} & = \sigma _{max} A \\ P_{max} & = \left( 200\ \text{MPa} \right)\left( 863.938\ \text{mm}^2 \right)\\ P_{max} & = \left( 200\ \frac{\text{N}}{\text{mm}^2} \right)\left( 863.938\ \text{mm}^2 \right)\\ P_{max} & = 172788\ \text{N} \\ P_{max} & = 172.8\ \text{kN}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}