Tag Archives: Mechanics

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*}

College Physics by Openstax Chapter 2 Problem 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 m/s2a=0.0500 \ \text{m/s}^2; t=8.00 minst=8.00 \ \text{mins}; and v0=4.00 m/sv_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula vf=v0+atv_f=v_0+at. We substitute the given values.

vf=v0+atvf=4.00 m/s+(0.0500 m/s2)(8.00 min×60 sec⁡1 min⁡)vf=28.0 m/s  (Answer)\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of tt, we have

t=vf−v0at=0 m/s−28 m/s−0.550 m/s2t=50.91 sec⁡  (Answer)\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), Δx \Delta x, or distance traveled is computed using the formula Δx=v0t+12at2 \Delta x=v_0 t+\frac{1}{2} at^2.

Δx=v0t+12at2Δx=(4.0 m/s)(480 s)+12(0.0500 m/s2)(480 s)2Δx=7680 m  (Answer)\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula Δx=vf2−v022a\Delta x=\frac{v_f^2-v_0^2}{2 a}.

Δx=(0 m/s)2−(28.0 m/s)22(−0.550 m/s2)Δx=712.73 m  (Answer)\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 1-4: The length of the American football field in meters

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PROBLEM:

American football is played on a 100-yard-long field, excluding the end zones. How long is the field in meters? (Assume that 1 meter equals 3.281 feet.)

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SOLUTION:

100 yard=100 yard×3 feet1 yard×1 m3.281 feet=91.4 m  (Answer)\begin{aligned} 100 \ \text{yard} & = 100 \ \bcancel{\text{yard}} \times \frac{3\ \bcancel{\text{feet}}}{1 \ \bcancel{\text{yard}}}\times \frac{1 \ \text{m}}{3.281\ \bcancel{\text{feet}}} \\ \\ & =91.4 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}
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Problem 1-2: Converting car speed of 33 m/s to kilometers per hour and determining if it exceeds the speed limit

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PROBLEM:

A car is traveling at a speed of 33 m/s.
(a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?

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SOLUTION:

Part A

33 m/s=33 ms×1 km1000 m×3600 s1 hr=118.8 km/hr  (Answer)\begin{aligned} 33 \ \text{m/s} & =33\ \frac{\text{m}}{\text{s}} \times \frac{1\ \text{km}}{1000 \ \text{m}} \times \frac{3600\ \text{s}}{1 \ \text{hr}} \\ \\ & =118.8 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{aligned}

Part B

At 118.8 km/h, the car is traveling faster than the speed limit of 90 km/h. (Answer)

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