Author Archives: Engineering Math

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*}

Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 3

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PROBLEM:

Evaluate lim⁡x→π4(tan⁡ x+sin⁡ x)\displaystyle \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right).

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SOLUTION:

lim⁡x→π4(tan⁡ x+sin⁡ x)=lim⁡x→π4(tan⁡ x)+lim⁡x→π4(sin⁡ x)=tan⁡ π4+sin⁡ π4=1+22=2+22  (Answer)\begin{align*} \lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x+\sin\:x\right) & =\lim\limits_{x\to \frac{\pi }{4}}\left(\tan\:x\right)+\lim\limits_{x\to \frac{\pi }{4}}\left(\sin\:x\right)\\ & =\tan\:\frac{\pi }{4}+\sin\:\frac{\pi }{4}\\ & =1+\frac{\sqrt{2}}{2}\\ & =\frac{2+\sqrt{2}}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 2

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PROBLEM:

Evaluate lim⁡x→3(4x+2x+4)\displaystyle \lim\limits_{x\to 3}\left(\frac{4x+2}{x+4}\right).

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SOLUTION:

lim⁡x→3(4x+2x+4)=lim⁡x→3(4x+2)lim⁡x→3(x+4)=lim⁡x→3(4x)+lim⁡x→3(2)lim⁡x→3(x)+lim⁡x→3(4)=4⋅lim⁡x→3(x)+23+4=4⋅3+23+4=12+27=147=2  (Answer)\begin{align*} \lim_{x\to 3}\left(\frac{4x+2}{x+4}\right)& =\frac{\lim\limits_{x\to 3}\left(4x+2\right)}{\lim\limits_{x\to 3}\left(x+4\right)}\\ & =\frac{\lim\limits_{x\to 3}\left(4x\right)+\lim\limits_{x\to 3}\left(2\right)}{\lim\limits_{x\to 3}\left(x\right)+\lim\limits_{x\to 3}\left(4\right)}\\ & =\frac{4\cdot \lim\limits_{x\to 3}\left(x\right)+2}{3+4}\\ & =\frac{4\cdot 3+2}{3+4}\\ & =\frac{12+2}{7}\\ & =\frac{14}{7}\\ & =2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 1

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PROBLEM:

Evaluate lim⁡x→2(x2−4x+3)\displaystyle \lim _{x\to 2}\left(x^2-4x+3\right).

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SOLUTION:

lim⁡x→2(x2−4x+3)=lim⁡x→2(x2)−lim⁡x→2(4x)+lim⁡x→2(3)=[lim⁡x→2(x)]2−4lim⁡x→2(x)+3=(2)2−4(2)+3=−1  (Answer)\begin{align*} \lim_{x\to 2}\left(x^2-4x+3\right)& = \lim_{x\to 2}\left(x^2\right)-\lim_{x\to 2}\left(4x\right)+\lim_{x\to 2}\left(3\right)\\ & =\left[\lim_{x\to 2}\left(x\right)\right]^2-4\lim_{x\to 2}\left(x\right)+3\\ & =\left(2\right)^2-4\left(2\right)+3\\ & =-1 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 10

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PROBLEM:

If  f(x)=4x+3\displaystyle f\left(x\right)=\frac{4}{x+3} and  g(x)=x2−3\displaystyle \:g\left(x\right)=x^2-3 , find f[g(x)]\displaystyle f\left[g\left(x\right)\right] and g[f(x)]\displaystyle g\left[f\left(x\right)\right].

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SOLUTION:

Part A

f[g(x)]=4(x2−3)+3=4x2  (Answer)\begin{align*} f\left[g\left(x\right)\right] & =\frac{4}{\left(x^2-3\right)+3}\\ & =\frac{4}{x^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

g[f(x)]=(4x+3)2−3=16(x+3)2−3=16−3(x+3)2(x+3)2=16−3(x2+6x+9)(x+3)2=16−3x2−18x−27(x+3)2=−3x2−18x−11(x+3)2  (Answer)\begin{align*} g\left[f\left(x\right)\right] & =\left(\frac{4}{x+3}\right)^2-3\\ & =\frac{16}{\left(x+3\right)^2}-3\\ & =\frac{16-3\left(x+3\right)^2}{\left(x+3\right)^2}\\ & =\frac{16-3\left(x^2+6x+9\right)}{\left(x+3\right)^2}\\ & =\frac{16-3x^2-18x-27}{\left(x+3\right)^2}\\ & =\frac{-3x^2-18x-11}{\left(x+3\right)^2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 9

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PROBLEM:

If f(x)=3x2−4x+1\displaystyle f\left(x\right)=3x^2-4x+1, find f(h+3)−f(3)h, h≠0\displaystyle \frac{f\left(h+3\right)-f\left(3\right)}{h},\:h\ne 0.

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SOLUTION:

f(h+3)−f(3)h=[3(h+3)2−4(h+3)+1]−[3(3)2−4(3)+1]h=3(h2+6h+9)−4h−12+1−16h=3h2+18h+27−4h−12+1−16h=3h2+14hh=h(3h+14)h=3h+14  (Answer)\begin{align*} \frac{f\left(h+3\right)-f\left(3\right)}{h} & =\frac{\left[3\left(h+3\right)^2-4\left(h+3\right)+1\right]-\left[3\left(3\right)^2-4\left(3\right)+1\right]}{h} \\ & =\frac{3\left(h^2+6h+9\right)-4h-12+1-16}{h}\\ & =\frac{3h^2+18h+27-4h-12+1-16}{h}\\ & =\frac{3h^2+14h}{h}\\ & =\frac{h\left(3h+14\right)}{h}\\ & =3h+14 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 8

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PROBLEM:

If f(x)=x2+1\displaystyle f\left(x\right)=x^2+1, find f(x+h)−f(x)h, h≠0\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h},\:h\ne 0.

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SOLUTION:

f(x+h)−f(x)h=[(x+h)2+1]−(x2+1) h=x2+2xh+h2+1−x2−1h=2xh+h2h=h(2x+h)h=2x+h  (Answer)\begin{align*} \displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h} & =\frac{\left[\left(x+h\right)^2+1\right]-\left(x^2+1\right)\:}{h}\\ \\ & =\frac{x^2+2xh+h^2+1-x^2-1}{h}\\ \\ & =\frac{2xh+h^2}{h}\\ \\ & =\frac{h\left(2x+h\right)}{h}\\ \\ & =2x+h \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1, Problem 7

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PROBLEM:

A right circular cylinder, a radius of base xx, height yy, is inscribed in a right circular cone, radius of base rr and a height hh. Express yy as a function of xx (rr and hh are constants).

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SOLUTION:

Refer to the figure below for an elevation view.

Schematic Diagram of a right circular cylinder inscribed in a right circular cone.
Diagram of a right circular cylinder with a base radius of r and height y inscribed in a right circular cone with base radius r and height h.

By ratio and proportion of two similar triangles, we have

yr−x=hry=h(r−x) r  (Answer)\begin{align*} \frac{y}{r-x} & = \frac{h}{r} \\ y & =\frac{h\left(r-x\right)\:}{r} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 6

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PROBLEM:

The stiffness of a beam of rectangular cross-section is proportional to the breadth and the cube of the depth. If the breadth is 20 cm, express the stiffness as a function of the depth.

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SOLUTION:

Let SS=stiffness, bb=breadth, and dd=depth

S=bd3S=20d3  (Answer)\begin{align*} S & =bd^3 \\ S & = 20 d^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Differential and Integral Calculus by Feliciano and Uy, Exercise 1.1 Problem 5

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PROBLEM:

Express the area AA of an equilateral triangle as a function of its side xx.

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SOLUTION:

From the formula of the area of a triangle, A=12absin⁡(θ)\displaystyle A=\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right). Also, we know that the interior angle of an equilateral triangle is 60 degrees, and sin⁡ 60∘=32\displaystyle \sin\:60^{\circ} =\frac{\sqrt{3}}{2}.

A=12absin⁡(θ)A=12⋅x⋅x⋅sin⁡ 60∘A=12⋅x2⋅32A=34x2  (Answer)\begin{align*} A & =\frac{1}{2} \text{a}\text{b} \sin\left(\theta \right) \\ A & =\frac{1}{2} \cdot x\cdot x\cdot \sin\:60^{\circ} \\ A & =\frac{1}{2}\cdot x^2\cdot \frac{\sqrt{3}}{2} \\ A & =\frac{\sqrt{3}}{4}x^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}
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