Mechanics of Materials Practice

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1.1Compute the Normal Stress in a Tension Member

Exam I | Problem 1.1 |

A steel bar carries a tensile force of $24\ \text{kN}$ and has a cross-sectional area of $600\ \text{mm}^2$.

What is the average normal stress in the bar?

Solution

Use the axial stress formula:

\sigma = \frac{P}{A}

Substitute the values:

\sigma = \frac{24\,000}{600} = 40\ \text{N/mm}^2

So the normal stress is

40\ \text{MPa}

1.2Find Engineering Strain from Elongation

Exam I | Problem 1.2 |

A bar with original length $750\ \text{mm}$ elongates by $1.5\ \text{mm}$.

What is the engineering strain?

Solution

Use the strain definition:

\varepsilon = \frac{\delta}{L}

Substitute the values:

\varepsilon = \frac{1.5}{750} = 0.002

So the engineering strain is

2.0 \times 10^{-3}

1.3Use Hooke's Law for a Linear Elastic Material

Exam I | Problem 1.3 |

Hooke's Law, Stress-Strain

A linearly elastic material has Young's modulus $E = 210\ \text{GPa}$. If it experiences a normal stress of $105\ \text{MPa}$, what is the axial strain?

Solution

Use Hooke's law:

\sigma = E\varepsilon

Solve for strain:

\varepsilon = \frac{\sigma}{E}

Substitute:

\varepsilon = \frac{105}{210\,000} = 0.0005

So the axial strain is

5.0 \times 10^{-4}

1.4Compute Free Thermal Expansion

Exam I | Problem 1.4 |

A steel rod has length $2.5\ \text{m}$ and coefficient of thermal expansion

\alpha = 12 \times 10^{-6}/^\circ\text{C}.

If the temperature increases by $20^\circ\text{C}$ and the rod is free to expand, how much does its length change?

Solution

Use free thermal expansion:

\delta_{th} = \alpha L \Delta T

Substitute:

\delta_{th} = (12 \times 10^{-6})(2.5)(20)

\delta_{th} = 6.0 \times 10^{-4}\ \text{m}

Convert to millimeters:

\delta_{th} = 0.6\ \text{mm}

1.5Find the Maximum Shear Stress in a Circular Shaft

Exam I | Problem 1.5 |

A solid circular shaft with diameter $60\ \text{mm}$ carries a torque of $600\ \text{N}\cdot\text{m}$.

What is the maximum shear stress in the shaft?

Solution

Use the torsion formula at the outer surface:

\tau_{max} = \frac{Tc}{J}

For a solid circular shaft,

J = \frac{\pi d^4}{32}, \quad c = \frac{d}{2}

Substitute $T = 600\,000\ \text{N}\cdot\text{mm}$ and $d = 60\ \text{mm}$:

\tau_{max} = \frac{600\,000(30)}{\pi(60^4)/32}

\tau_{max} \approx 14.1\ \text{MPa}

1.6Compute the Angle of Twist

Exam I | Problem 1.6 |

A solid circular shaft has length $1.5\ \text{m}$, diameter $50\ \text{mm}$, shear modulus $G = 80\ \text{GPa}$, and applied torque $900\ \text{N}\cdot\text{m}$.

What is the angle of twist?

Solution

Use the torsion deformation formula:

\phi = \frac{TL}{JG}

For a solid circular shaft,

J = \frac{\pi d^4}{32}

Substitute in consistent units:

T = 900\,000\ \text{N}\cdot\text{mm}, \quad L = 1500\ \text{mm}, \quad G = 80\,000\ \text{N/mm}^2

Then

\phi = \frac{900\,000(1500)}{\left[\pi(50^4)/32\right](80\,000)}

\phi \approx 0.0275\ \text{rad}

In degrees, this is about

1.6^\circ

1.7Find the Maximum Bending Stress in a Rectangle

Exam I | Problem 1.7 |

A beam section is rectangular with width $100\ \text{mm}$ and height $150\ \text{mm}$. The internal bending moment at a section is $8\ \text{kN}\cdot\text{m}$.

What is the maximum bending stress?

Solution

Use the flexure formula:

\sigma_{max} = \frac{Mc}{I}

For a rectangle,

I = \frac{bh^3}{12}

Here,

I = \frac{100(150^3)}{12} = 28\,125\,000\ \text{mm}^4

and

c = \frac{150}{2} = 75\ \text{mm}

Convert the moment:

M = 8\ \text{kN}\cdot\text{m} = 8\,000\,000\ \text{N}\cdot\text{mm}

Now compute:

\sigma_{max} = \frac{8\,000\,000(75)}{28\,125\,000} \approx 21.3\ \text{MPa}

1.8Compute Maximum Beam Shear Stress in a Rectangle

Exam I | Problem 1.8 |

A rectangular beam has width $50\ \text{mm}$ and height $150\ \text{mm}$. The internal shear force is $15\ \text{kN}$.

What is the maximum shear stress at the neutral axis?

Solution

For a rectangular section, the maximum shear stress is

\tau_{max} = \frac{3V}{2A}

where

A = bh = 50(150) = 7500\ \text{mm}^2

Substitute:

\tau_{max} = \frac{3(15\,000)}{2(7500)} = 3.0\ \text{MPa}

So the maximum shear stress is

3.0\ \text{MPa}

1.9Interpret a Concentrated Load in a Shear Diagram

Exam I | Problem 1.9 |

A beam's shear-force diagram is being drawn from left to right. At one point, the beam crosses a $12\ \text{kN}$ downward concentrated load.

What happens to the shear-force diagram at that point?

Solution

A downward concentrated load causes a downward jump in the shear diagram by the same magnitude.

So the shear-force diagram drops by

12\ \text{kN}

at that point.

1.10Find the Euler Buckling Load

Exam I | Problem 1.10 |

A pinned-pinned column has

E = 200\ \text{GPa}, \quad I = 1.0 \times 10^6\ \text{mm}^4, \quad L = 3\ \text{m}.

What is the critical buckling load?

Solution

Use Euler buckling with $K = 1$ for pinned-pinned ends:

P_{cr} = \frac{\pi^2EI}{(KL)^2}

Substitute in consistent units:

E = 200\,000\ \text{N/mm}^2, \quad L = 3000\ \text{mm}

P_{cr} = \frac{\pi^2(200\,000)(1.0 \times 10^6)}{3000^2}

P_{cr} \approx 219\ \text{kN}

Difficulty

Integrated

2.1Sum the Elongation of a Two-Segment Bar

Exam II | Problem 2.1 |

A steel bar has two segments in series. Segment 1 has length $400\ \text{mm}$ and area $200\ \text{mm}^2$. Segment 2 has length $600\ \text{mm}$ and area $300\ \text{mm}^2$. The bar carries a tensile force of $12\ \text{kN}$ throughout, and $E = 200\ \text{GPa}$.

What is the total elongation?

Solution

Use the deformation formula on each segment:

\delta = \sum \frac{PL}{AE}

For segment 1:

\delta_1 = \frac{12\,000(400)}{200(200\,000)} = 0.12\ \text{mm}

For segment 2:

\delta_2 = \frac{12\,000(600)}{300(200\,000)} = 0.12\ \text{mm}

So the total elongation is

\delta_{tot} = 0.24\ \text{mm}

2.2Analyze a Stepped Bar with Constant Axial Force

Exam II | Problem 2.2 |

A stepped steel bar is in tension with an axial force of $18\ \text{kN}$. Its segments are:

(500\ \text{mm}, 250\ \text{mm}^2),\quad (300\ \text{mm}, 500\ \text{mm}^2),\quad (700\ \text{mm}, 250\ \text{mm}^2)

Take $E = 200\ \text{GPa}$.

Find the stress in each segment and the total elongation.

Solution

The axial force is the same in every segment, so the stress in each segment is

\sigma = \frac{P}{A}

Segment 1:

\sigma_1 = \frac{18\,000}{250} = 72\ \text{MPa}

Segment 2:

\sigma_2 = \frac{18\,000}{500} = 36\ \text{MPa}

Segment 3:

\sigma_3 = \frac{18\,000}{250} = 72\ \text{MPa}

Now compute elongation segment by segment:

\delta_{tot} = \sum \frac{PL}{AE}

\delta_{tot} = \frac{18\,000(500)}{250(200\,000)} +\frac{18\,000(300)}{500(200\,000)} +\frac{18\,000(700)}{250(200\,000)}

\delta_{tot} = 0.36 + 0.054 + 0.252 = 0.486\ \text{mm}

2.3Find the Stress from a Fully Restrained Temperature Rise

Exam II | Problem 2.3 |

A steel bar with length $2\ \text{m}$ is fixed between rigid walls. Its coefficient of thermal expansion is

\alpha = 12 \times 10^{-6}/^\circ\text{C},

and $E = 200\ \text{GPa}$. If the temperature rises by $35^\circ\text{C}$, what thermal stress develops?

Solution

For a fully restrained bar, the total strain is zero, so the thermal stress is

\sigma = E\alpha\Delta T

Substitute:

\sigma = (200\,000)(12 \times 10^{-6})(35)

\sigma = 84\ \text{MPa}

Because the bar is prevented from expanding, the stress is compressive.

2.4Share Load Between Parallel Bars

Exam II | Problem 2.4 |

Two bars connect rigid plates in parallel. Each bar is $800\ \text{mm}$ long and has cross-sectional area $400\ \text{mm}^2$. One bar is steel with $E = 200\ \text{GPa}$, and the other is aluminum with $E = 70\ \text{GPa}$. The plates are pulled by a total tensile load of $27\ \text{kN}$.

How much force does each bar carry?

Solution

Because the bars are in parallel between rigid plates, they have the same elongation. So the force in each bar is proportional to its stiffness:

k = \frac{AE}{L}

Since $A$ and $L$ are the same, the force ratio is

$$ F_s : F_a = E_s : E_a = 200 : 70 = 20 : 7 $$

The total load is $27\ \text{kN}$, so there are $27$ parts total.

Steel bar:

F_s = 27 \cdot \frac{20}{27} = 20\ \text{kN}

Aluminum bar:

F_a = 27 \cdot \frac{7}{27} = 7\ \text{kN}

The common elongation is

\delta = \frac{F_sL}{AE_s} = \frac{20\,000(800)}{400(200\,000)} = 0.2\ \text{mm}

2.5Find Principal Stresses from a Plane Stress State

Exam II | Problem 2.5 |

A point in a member is under the plane stress state

\sigma_x = 84\ \text{MPa}, \quad \sigma_y = 36\ \text{MPa}, \quad \tau_{xy} = 18\ \text{MPa}.

Find the principal stresses and the maximum in-plane shear stress.

Solution

Use the plane stress transformation results:

\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{ \left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2 }

Compute the average stress:

\frac{84+36}{2} = 60\ \text{MPa}

and the radius:

\sqrt{24^2 + 18^2} = \sqrt{900} = 30\ \text{MPa}

So the principal stresses are

\sigma_1 = 90\ \text{MPa}, \quad \sigma_2 = 30\ \text{MPa}

The maximum in-plane shear stress is

\tau_{max} = 30\ \text{MPa}

2.6Transform Stress to a Rotated Plane

Exam II | Problem 2.6 |

Using the same plane stress state

\sigma_x = 84\ \text{MPa}, \quad \sigma_y = 36\ \text{MPa}, \quad \tau_{xy} = 18\ \text{MPa},

find the normal stress and shear stress on a plane rotated $45^\circ$ counterclockwise.

Solution

Use the stress transformation equations:

\sigma_{x'} = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta

\tau_{x'y'} = -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta

With $\theta = 45^\circ$, we have

\cos 90^\circ = 0, \quad \sin 90^\circ = 1

\sigma_{x'} = 60 + 18 = 78\ \text{MPa}

and

\tau_{x'y'} = -24\ \text{MPa}

The normal stress on the rotated plane is $78\ \text{MPa}$, and the shear stress is $-24\ \text{MPa}$.

2.7Use the Beam Shear Formula on a Rectangular Section

Exam II | Problem 2.7 |

A rectangular beam has width $50\ \text{mm}$ and height $150\ \text{mm}$. The internal shear force is $15\ \text{kN}$.

Find the maximum shear stress at the neutral axis using the beam shear formula.

Solution

For a rectangular section, the maximum shear stress occurs at the neutral axis and can be found from

\tau = \frac{VQ}{It}

Here,

I = \frac{bh^3}{12} = \frac{50(150^3)}{12}

At the neutral axis, the area above the cut is

A' = b\left(\frac{h}{2}\right) = 50(75)

and its centroid is $37.5\ \text{mm}$ from the neutral axis, so

Q = A'\bar{y} = 50(75)(37.5)

Using $t = b = 50\ \text{mm}$, the result simplifies to

\tau_{max} = \frac{3V}{2A}

with

A = bh = 50(150) = 7500\ \text{mm}^2

\tau_{max} = \frac{3(15\,000)}{2(7500)} = 3.0\ \text{MPa}

Difficulty

Applied

3.1Integrate the Deflection of a Cantilever

Final | Problem 3.1 |

A cantilever beam has length $2\ \text{m}$ and an end load of $4\ \text{kN}$. The beam has $E = 200\ \text{GPa}$ and $I = 8.0 \times 10^6\ \text{mm}^4$.

Find the tip deflection using the moment-curvature relation.

Solution

Take $x$ from the fixed end. The bending moment is

$$ M(x) = -P(L-x) $$

and the beam equation is

EI\frac{d^2v}{dx^2} = M(x)

EI\frac{d^2v}{dx^2} = -P(L-x)

Integrate once:

EI\frac{dv}{dx} = -PLx + \frac{Px^2}{2} + C_1

At the fixed end, the slope is zero, so $C_1 = 0$.

Integrate again:

EIv = -\frac{PLx^2}{2} + \frac{Px^3}{6} + C_2

At the fixed end, the deflection is zero, so $C_2 = 0$.

At $x = L$,

v(L) = -\frac{PL^3}{3EI}

Substitute $P = 4000\ \text{N}$, $L = 2000\ \text{mm}$, $E = 200\,000\ \text{N/mm}^2$, and $I = 8.0 \times 10^6\ \text{mm}^4$:

|v(L)| = \frac{4000(2000^3)}{3(200\,000)(8.0 \times 10^6)} \approx 6.7\ \text{mm}

So the tip deflection is about $6.7\ \text{mm}$ downward.

3.2Convert Power to Shaft Diameter

Final | Problem 3.2 |

A motor delivers $12\ \text{kW}$ at $900\ \text{rpm}$ to a solid circular shaft. If the allowable shear stress is $40\ \text{MPa}$, what minimum shaft diameter is required?

Solution

First convert power and speed to torque:

P = T\omega

The angular speed is

\omega = 2\pi\left(\frac{900}{60}\right) = 30\pi\ \text{rad/s}

So the torque is

T = \frac{12\,000}{30\pi} \approx 127.3\ \text{N}\cdot\text{m}

Now use the torsion stress formula for a solid shaft:

\tau_{max} = \frac{16T}{\pi d^3}

Solve for $d$:

d = \left(\frac{16T}{\pi\tau_{allow}}\right)^{1/3}

Substitute $T = 127\,300\ \text{N}\cdot\text{mm}$ and $\tau_{allow} = 40\ \text{N/mm}^2$:

d \approx 25.4\ \text{mm}

So the minimum practical diameter is

26\ \text{mm}

3.3Combine Axial Load and Bending

Final | Problem 3.3 |

A member carries a compressive axial load of $30\ \text{kN}$ and an internal bending moment of $6\ \text{kN}\cdot\text{m}$. Its cross-section is a rectangle with width $80\ \text{mm}$ and height $160\ \text{mm}$.

Find the normal stress at the top fiber and at the bottom fiber.

Solution

First compute the axial stress:

\sigma_{ax} = \frac{P}{A} = \frac{30\,000}{80(160)} = 2.34\ \text{MPa}

This is compressive.

Next compute the bending stress:

\sigma_b = \frac{Mc}{I}

For a rectangle,

I = \frac{bh^3}{12} = \frac{80(160^3)}{12} = 34\,133\,333\ \text{mm}^4

and

c = 80\ \text{mm}

Convert the moment:

M = 6\ \text{kN}\cdot\text{m} = 6\,000\,000\ \text{N}\cdot\text{mm}

\sigma_b = \frac{6\,000\,000(80)}{34\,133\,333} \approx 14.1\ \text{MPa}

If the moment puts the top fiber in compression, then:

\sigma_{top} = 2.34 + 14.1 = 16.4\ \text{MPa}

\sigma_{bottom} = 14.1 - 2.34 = 11.8\ \text{MPa}

So the top fiber is more highly compressed, and the bottom fiber is in tension.

3.4Decide Which Column Failure Mode Governs

Final | Problem 3.4 |

A steel column has length $3\ \text{m}$, pinned-pinned ends, Young's modulus $E = 200\ \text{GPa}$, and a circular cross-section with diameter $40\ \text{mm}$. The yield strength is $250\ \text{MPa}$.

Compute the Euler critical load and the yield load. Which one is smaller?

Solution

First compute the second moment of area for a solid circle:

I = \frac{\pi d^4}{64}

With $d = 40\ \text{mm}$,

I = \frac{\pi(40^4)}{64} \approx 1.26 \times 10^5\ \text{mm}^4

Euler buckling load:

P_{cr} = \frac{\pi^2EI}{L^2}

Substitute:

P_{cr} \approx \frac{\pi^2(200\,000)(1.26 \times 10^5)}{3000^2} \approx 27.6\ \text{kN}

Now compute the yield load:

A = \frac{\pi d^2}{4} \approx 1257\ \text{mm}^2

P_y = \sigma_y A = 250(1257) \approx 314\ \text{kN}

Since $27.6\ \text{kN} < 314\ \text{kN}$, Euler buckling governs.

3.5Check a Plane Stress State with von Mises

Final | Problem 3.5 |

A ductile part has the plane stress state

\sigma_x = 70\ \text{MPa}, \quad \sigma_y = 10\ \text{MPa}, \quad \tau_{xy} = 20\ \text{MPa}.

If the yield strength is $75\ \text{MPa}$, find the von Mises stress and decide whether yielding is predicted.

Solution

Use the plane-stress von Mises formula:

\sigma_{vm} = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2}

Substitute:

\sigma_{vm} = \sqrt{70^2 - 70(10) + 10^2 + 3(20^2)}

\sigma_{vm} = \sqrt{4900 - 700 + 100 + 1200}

\sigma_{vm} = \sqrt{5500} \approx 74.2\ \text{MPa}

Because $74.2\ \text{MPa} < 75\ \text{MPa}$, yielding is not predicted. The part is just barely safe.

Difficulty

Challenge

4.1Include a Stress Concentration in a Torsion Design Check

Final | Problem 4.1 |

A solid circular shaft has diameter $50\ \text{mm}$ and a shoulder with stress concentration factor $K_t = 1.6$. If the allowable maximum shear stress is $50\ \text{MPa}$, what is the largest torque the shaft can carry?

Solution

The local maximum shear stress is

\tau_{max} = K_t \tau_{nom}

So the nominal stress must satisfy

\tau_{nom} \le \frac{50}{1.6} = 31.25\ \text{MPa}

For a solid circular shaft,

\tau_{nom} = \frac{16T}{\pi d^3}

Solve for $T$:

T = \frac{\tau_{nom}\pi d^3}{16}

Substitute $d = 50\ \text{mm}$:

T = \frac{31.25\pi(50^3)}{16} \approx 7.67 \times 10^5\ \text{N}\cdot\text{mm}

So the largest torque is

767\ \text{N}\cdot\text{m}

4.2Solve a Thermal-Indeterminate Parallel-Bar System

Final | Problem 4.2 |

Two bars connect rigid plates in parallel. Each bar is $800\ \text{mm}$ long and has cross-sectional area $400\ \text{mm}^2$. One bar is steel with $E = 200\ \text{GPa}$ and $\alpha = 12 \times 10^{-6}/^\circ\text{C}$. The other is aluminum with $E = 70\ \text{GPa}$ and $\alpha = 23 \times 10^{-6}/^\circ\text{C}$.

The temperature rises by $10^\circ\text{C}$, and the plates are also pulled by a total tensile load of $18\ \text{kN}$.

Find the force in each bar.

Solution

Because the plates are rigid, both bars have the same total elongation. Let $\delta$ be that common elongation.

For each bar,

F_i = k_i(\delta - \delta_{th,i}), \quad k_i = \frac{AE_i}{L}

Compute the stiffnesses:

k_s = \frac{400(200\,000)}{800} = 100\,000\ \text{N/mm}

k_a = \frac{400(70\,000)}{800} = 35\,000\ \text{N/mm}

Compute the free thermal expansions:

\delta_{th,s} = (12 \times 10^{-6})(800)(10) = 0.12\ \text{mm}

\delta_{th,a} = (23 \times 10^{-6})(800)(10) = 0.23\ \text{mm}

Force equilibrium gives

F_s + F_a = 18\,000

and

100\,000(\delta - 0.12) + 35\,000(\delta - 0.23) = 18\,000

135\,000\delta = 38\,050

\delta \approx 0.282\ \text{mm}

Now compute the forces:

F_s = 100\,000(0.282 - 0.12) \approx 16.2\ \text{kN}

F_a = 35\,000(0.282 - 0.23) \approx 1.8\ \text{kN}

So the steel bar carries about $16.2\ \text{kN}$ and the aluminum bar carries about $1.8\ \text{kN}$.

4.3Check a Shaft in Combined Bending and Torsion

Final | Problem 4.3 |

A solid circular shaft with diameter $60\ \text{mm}$ carries a bending moment of $300\ \text{N}\cdot\text{m}$ and a torque of $500\ \text{N}\cdot\text{m}$.

Find the principal stresses at the outer surface and the von Mises stress. If the yield strength is $40\ \text{MPa}$, is the shaft safe?

Solution

At the outer surface, the bending stress is

\sigma = \frac{32M}{\pi d^3}

and the torsional shear stress is

\tau = \frac{16T}{\pi d^3}

Substitute $d = 60\ \text{mm}$, $M = 300\,000\ \text{N}\cdot\text{mm}$, and $T = 500\,000\ \text{N}\cdot\text{mm}$:

\sigma \approx 14.1\ \text{MPa}, \quad \tau \approx 11.8\ \text{MPa}

Now treat this as a plane stress state with $\sigma_x = 14.1\ \text{MPa}$, $\sigma_y = 0$, and $\tau_{xy} = 11.8\ \text{MPa}$.

The principal stresses are

\sigma_{1,2} = \frac{\sigma_x}{2} \pm \sqrt{\left(\frac{\sigma_x}{2}\right)^2 + \tau_{xy}^2}

\sigma_1 \approx 20.8\ \text{MPa}, \quad \sigma_2 \approx -6.7\ \text{MPa}

The von Mises stress is

\sigma_{vm} = \sqrt{\sigma_x^2 + 3\tau_{xy}^2}

\sigma_{vm} \approx \sqrt{14.1^2 + 3(11.8^2)} \approx 24.8\ \text{MPa}

Since $24.8\ \text{MPa} < 40\ \text{MPa}$, the shaft is safe by the von Mises criterion.

4.4Apply Goodman Fatigue Design

Final | Problem 4.4 |

A member has alternating stress and mean stress

\sigma_a = 40\ \text{MPa}, \quad \sigma_m = 100\ \text{MPa}.

The endurance limit is $S_e = 240\ \text{MPa}$ and the ultimate tensile strength is $S_{ut} = 600\ \text{MPa}$.

What factor of safety does the Goodman relation predict?

Solution

Use the Goodman design relation:

\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} \le \frac{1}{n}

Substitute the values:

\frac{40}{240} + \frac{100}{600} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}

\frac{1}{n} = \frac{1}{3}

which gives

$$ n = 3 $$

4.5Check an Eccentrically Loaded Column

Final | Problem 4.5 |

A solid steel column has diameter $30\ \text{mm}$, length $3\ \text{m}$, and modulus $E = 200\ \text{GPa}$. It carries a compressive load of $8\ \text{kN}$ applied with eccentricity $15\ \text{mm}$.

Find the maximum compressive stress and the Euler buckling load. If the allowable compressive stress is $70\ \text{MPa}$, which check is more restrictive?

Solution

The eccentric load creates both axial stress and bending stress.

Area:

A = \frac{\pi d^2}{4} = \frac{\pi(30^2)}{4} \approx 706.9\ \text{mm}^2

Axial stress:

\sigma_{ax} = \frac{P}{A} = \frac{8000}{706.9} \approx 11.3\ \text{MPa}

Bending moment from eccentricity:

M = Pe = 8000(15) = 120\,000\ \text{N}\cdot\text{mm}

For a solid circular section,

I = \frac{\pi d^4}{64} \approx 39\,789\ \text{mm}^4

and

c = 15\ \text{mm}

Bending stress:

\sigma_b = \frac{Mc}{I} = \frac{120\,000(15)}{39\,789} \approx 45.3\ \text{MPa}

So the maximum compressive stress is

\sigma_{max} \approx 11.3 + 45.3 = 56.6\ \text{MPa}

Now compute Euler buckling:

P_{cr} = \frac{\pi^2EI}{L^2}

with $L = 3000\ \text{mm}$:

P_{cr} \approx \frac{\pi^2(200\,000)(39\,789)}{3000^2} \approx 8.7\ \text{kN}

Since the applied load is $8\ \text{kN}$, the column is below the Euler load but only by a small margin.

The compressive stress check is also below the $70\ \text{MPa}$ allowable. The more restrictive check is buckling, because $8\ \text{kN}$ is close to $8.7\ \text{kN}$.

Mechanics of Materials Practice

Selected problems

General shear formula

Direct

1.1Compute the Normal Stress in a Tension Member

1.2Find Engineering Strain from Elongation

1.3Use Hooke's Law for a Linear Elastic Material

1.4Compute Free Thermal Expansion

1.5Find the Maximum Shear Stress in a Circular Shaft

1.6Compute the Angle of Twist

1.7Find the Maximum Bending Stress in a Rectangle

1.8Compute Maximum Beam Shear Stress in a Rectangle

1.9Interpret a Concentrated Load in a Shear Diagram

1.10Find the Euler Buckling Load

Integrated

2.1Sum the Elongation of a Two-Segment Bar

2.2Analyze a Stepped Bar with Constant Axial Force

2.3Find the Stress from a Fully Restrained Temperature Rise

2.4Share Load Between Parallel Bars

2.5Find Principal Stresses from a Plane Stress State

2.6Transform Stress to a Rotated Plane

2.7Use the Beam Shear Formula on a Rectangular Section

Applied

3.1Integrate the Deflection of a Cantilever

3.2Convert Power to Shaft Diameter

3.3Combine Axial Load and Bending

3.4Decide Which Column Failure Mode Governs

3.5Check a Plane Stress State with von Mises

Challenge

4.1Include a Stress Concentration in a Torsion Design Check

4.2Solve a Thermal-Indeterminate Parallel-Bar System

4.3Check a Shaft in Combined Bending and Torsion

4.4Apply Goodman Fatigue Design

4.5Check an Eccentrically Loaded Column