Mechanics of materials/Problem set 6

Problem 6.1 (Problem 4.101 in Beer, 2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement

A bent beam fixed to a wall is subjected to force P

The bar is fixed to the wall and a force P is applied at point D on the bar. Determine the stress at point A and at point B.

Given

$P=8kN$

Solution

Step One: Draw Free Body Diagrams

A cross section of the beam at AB

Free body diagram at AB

Step Two: Calculate the cross sectional area

$A=b\times h$

$A=30\times 24$

$A=720m{m}^2$

Step Three: Calculate Inertia

$I_x=\frac{b\times d^3}{12}$

$I_x=\frac{30\times 24^3}{12}$

Step Four: Calculate the centroid

$c=\frac{h}{2}$

$c=\frac{24}{2}$

Step 5: Finding the eccentricity

$e=45-\frac{24}{2}$

Step 6: Calculating the bending moment

$M=P\times e$

$M=8\times ,033$

$M=264N\cdot m$

Step 7: Calculating stress at A

The stress at A can be separated into two parts, the stress due to centric loading and the stress due to bending

${\sigma }_A={\sigma }_{centric,A}+{\sigma }_{bending,A}$

${\sigma }_{centric,A}=\frac{-P}{A}=\frac{8\times 10^3}{7.2\times 10^{-}4}$

${\sigma }_{centric,A}=-11.11MPa$

The stress due to bending at point A can be represented as,

${\sigma }_{bending,A}=\frac{M{c}_A}{I}$

${\sigma }_{bending,A}=\frac{(264N\cdot m)(.012m)}{3.456\times (10^{-8})m^4}$

${\sigma }_{bending,A}=-91.67MPa$

Now adding the centric and bending stresses we receive,

${\sigma }_A=(-11.11MPa)-(91.67MPa)=-102.8MPa$

Step 8: Calculate the stress at point B

${\sigma }_B={\sigma }_{centric,B}+{\sigma }_{bending,B}$

${\sigma }_{centric,B}=\frac{-P}{A}=\frac{8\times 10^3}{7.2\times 10^{-}4}$

${\sigma }_{centric,B}=-11.11MPa$
The stress due to bending at point B can be represented as,

${\sigma }_{bending,B}=\frac{M{c}_B}{I}$

${\sigma }_{bending,B}=\frac{(264N\cdot m)(.012m)}{3.456\times (10^{-8})m^4}$

${\sigma }_{bending,B}=91.67MPa$

So, the total stress at B is

${\sigma }_B=(-11.11MPa)+(91.67MPa)=80.56MPa$

Problem 6.2 (Problem 4.103 in Beer, 2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement

The vertical portion of the press shown consists of a rectangular tube of wall thickness t=8mm. Knowing that the press has been tightened on wooden planks being glued together until P=20kN, determine the stress at (a) point A, (b) point B.

Problem 4.103

Solution

Step One: Draw Free Body Diagrams

Free body diagram

Step Two: Find Area of Cross section

Area of the cross section

Cross sectional area

$A=A_{outer}-A_{inner}$

(6.2-1)

$A=(60*80)-[(60-2(8))*(80-2(8))]$

(6.2-2)

$A=1984m{m}^2$

(6.2-3)

$A=1984*10^{-6}{m}^2$

(6.2-4)

Step 3: Calculate moment of Inertia

Moment of Inertia

Moment of Inertia

$I=I_{outer}-I_{inner}$

(6.2-5)

$I_{outer}=\frac{b{d}^3}{12}$

(6.2-6)

$I_{outer}=\frac{60*80^3}{12}$

(6.2-7)

$I_{outer}=256*10^{4}m{m}^4$

(6.2-8)

$I_{outer}=256*10^{-8}{m}^4$

(6.2-9)

Inner cross sectional area

$I_{inner}=\frac{b{d}^3}{12}$

(6.2-10)

$I_{inner}=\frac{40*60^3}{12}$

(6.2-11)

$I_{inner}=720000m{m}^4$

(6.2-12)

$I_{inner}=72*10^{-8}{m}^4$

(6.2-13)

$I=(256*10^{-8})-(72*10^{-8})$

(6.2-14)

$I=1.84*10^{-6}{m}^4$

(6.2-15)

Step 4: Find the moment and eccentricity

The internal forces in the cross section are equivalent to a centric force P and a bending couple M.

$c=\frac{80}{2}$

(6.2-16)

$c=40mm$

(6.2-17)

Moment M=P*eccentricity

$M=P*e$

(6.2-18)

Calculating the eccentricity

$e=200+\frac{80}{2}$

(6.2-19)

$e=240mm=0.240m$

(6.2-20)

Calculating the Moment

$M=(20*10^3)*(0.240)$

(6.2-21)

$M=4800N*m$

(6.2-22)

Step Five: Find the Stress at point A and point B

The stress is calculated by adding the stress due to centric force and stress due to the couple M.

${\sigma }_A={\sigma }_{centric}+{\sigma }_{bending}$

(6.2-23)

Part a: Stress at point A

${\sigma }_{centric}=\frac{P}{A}$

(6.2-24)

${\sigma }_{centric}=\frac{20*10^3}{1984*10^{-6}}$

(6.2-25)

${\sigma }_{centric}=10.08MPa$

(6.2-26)

${\sigma }_{bending}=\frac{M*c}{I}$

(6.2-27)

${\sigma }_{bending}=\frac{4800*0.04}{1.84*10^{-6}}$

(6.2-28)

${\sigma }_{bending}=104.35MPa$

(6.2-29)

${\sigma }_A=10.08+104.35$

(6.2-30)

${\sigma }_A=114.43MPa$

Part b: Stress at point B

${\sigma }_B={\sigma }_{centric}+{\sigma }_{bending}$

(6.2-31)

The centric stress force is equal to the same as the one derived in equation (6.2-26)

${\sigma }_{centric}=10.08MPa$

(6.2-32)

The bending stress force is equal to the same as the one derived in equation (6.2-29) but the negative.

${\sigma }_{bending}=-104.35MPa$

(6.2-33)

${\sigma }_B=10.08+(-104.35)$

(6.2-34)

${\sigma }_B=-94.27MPa$

Problem 6.3 (Problem 4.112 in Beer, 2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement

File:P4.11112.png

A metal tube with an outer diameter of 0.75 in. and an wall thickness of 0.08 in. is shown in Figure 6.3.1. Determine the largest offset that can be used if the maximum stress after the offset is introduced does not exceed four times the stress in the tube when it is straight.

Given

Outer diameter, $d_o=0.75in.$
Wall thickness, $t=0.08in.$

Solution

Step One: Draw free body diagrams

File:P4.112FBD.png

File:Fbd1 pb 112.png

Step Two: Solve for the area and centroidal moment of inertia

The area of the pipe is given by Equation 6.3-1

$A=\frac{\pi }{4}(d_o^2-d_i^2)=\frac{\pi }{4}(d_o^2-(d_o-2t{)}^2)$

(6.3-1)

and is calculated to be

$A=\frac{\pi }{4}(0.75^2[in.{]}^2-(0.75[in.]-2\times 0.08[in]{)}^2)=0.168in{.}^2$

The centroidal moment of inertia is given by Equation 6.3-2

$I=\frac{\pi }{64}(d_o^4-d_i^4)=\frac{\pi }{64}(d_o^4-(d_o-2t{)}^4)$

(6.3-2)

and is calculated to be

$I=\frac{\pi }{64}(0.75^4[in.{]}^4-(0.75[in.]-2\times 0.08[in]{)}^4)=9.58\times 10^{-3}in{.}^4$

Step Three: Formulate expressions for stress with and without offset

Without the offset in the member, the stress, ${\sigma }_c$ is calculated as if there is a centric loading, where $P$ is pressure and $A$ is area.

${\sigma }_c=\frac{P}{A}$

(6.3-3)

With the offset in the member, the stress, ${\sigma }_o$, $h$ is the offset, $c$ is the radius, and $I$ is the moment of inertia.

${\sigma }_o=\frac{P}{A}+\frac{Phc}{I}$

(6.3-4)

Step Four: Solve for the offset

The problem statement stipulates that the stress with the offset cannot exceed four times the stress without it. This can be used to solve for $h$.

${\sigma }_o=4{\sigma }_c$

(6.3-5)

$\frac{P}{A}+\frac{Phc}{I}=4\frac{P}{A}$

(6.3-6)

$\frac{Phc}{I}=3\frac{P}{A}$

(6.3-7)

Therefore $h$ can be solved for as,

$h=\frac{3I}{CA}$

(6.3-8)

Inserting the previously calculated values,

$h=\frac{3\times 9.58\times 10^{-3}[in{.}^4]}{\frac{0.75}{2}[in.]\times 0.168[in{.}^2]}=0.456in.$

Problem 6.4 (Problem 4.114 in Beer, 2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement

vertical rod attached to a cast iron hanger

A vertical rod is attached to the cast iron hanger at point A, as shown. Given the maximum allowable stresses in the hanger, determine the largest downward force and the largest upward force that can be exerted by the rod.

Given

Maximum allowable stresses are
${\sigma }_{all}=+5ksi$

${\sigma }_{all}=-12ksi$

Solution

Step One: Draw Free Body Diagrams

Free body diagram

Finding the area of the three sections

$A_1=(3in)(.75in)=2.25i{n}^2$

(6.4-1)

$A_2=(3in)(1.0in)=3i{n}^2$

(6.4-2)

$A_3=(3in)(.75in)=2.25i{n}^2$

(6.4-3)

Total area of the sections

$A_1+A_2+A_3=A=7.5i{n}^2$

(6.4-4)

Distance of the centroid of part 1 from left edge

$\overline{y_1}=(1in)+\frac{3in}{2}=2.5in$

(6.4-5)

Distance of the centroid of part 2 from left edge

$\overline{y_2}=\frac{1in}{2}=0.5in$

(6.4-6)

Distance of the centroid of part 3 from left edge

$\overline{y_3}=(1in)\frac{3in}{2}=2.5in$

(6.4-7)

$\overline{y}=\frac{A_1{y}_1+A_2{y}_2+A_3{y}_3}{A}=1.7in.$

(6.4-8)

Moment of Inertia

Using the parallel axis theorem, the moment of inertia about any point can be represented as,

$I=\frac{b{d}^3}{12}+A{d}^2$

(6.4-9)

The moment of inertia of part 1 about the centroid is

$I_1=\frac{b_1{d}_1^3}{12}+A_1{d}_1^2=\frac{b_1{d}_1^3}{12}+A_1({\overline{y}}_1-\overline{y}{)}^2$

(6.4-10)

The moment of inertia of part 2 about the centroid is

$I_2=\frac{b_2{d}_2^3}{12}+A_2{d}_2^2=\frac{b_2{d}_2^3}{12}+A_2({\overline{y}}_2-\overline{y}{)}^2$

(6.4-11)

Since I₁ and I₃ are similar and symmetric,

$I_3=I_1$

(6.4-12)

Now, the total moment of inertia can be found by adding all of the individual parts,

$I=I_1+I_2+I_3=(3.1275i{n}^4)+(4.57i{n}^4)+(3.1275i{n}^4)=10.825i{n}^4$

(6.4-13)

Calculating maximum force P

Let P be the maximum force that can be applied.
The bending moment due to force P at the centroid is

$M=Pd$

(6.4-14)

where d is the distance from the application point to the centroid, $d=3.2in.$
The total stress acting at point A is a combination of normal stress and bending stress

${\sigma }_x=-\frac{P}{A}-\frac{My}{I}$

(6.4-15)

From this point, knowing that a downward force at A would induce a stress such that the maximum stress occurs at the right side of the cross section, and when and upward force is applied at A, the maximum stress occurs at the left side of the cross section will allow us to solve for P. To do this, we must substitute both values for ${\sigma }_{all}$, +5ksi and -12ksi and take the smallest resulting P in both cases.
The maximum allowable downward force is found to be

$P_{max}=7.86kips$

The maximum allowable upward force is found to be

$P_{max}=9.15kips$

Problem 6.5 (Problem 4.115 in Beer, 2012)

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem Statement

A vertical rod is attached at point B to a cast iron hanger. Knowing that the allowable stresses in the hanger are ${\sigma }_{all}=+5ksi$ and ${\sigma }_{all}=-12ksi$, determine the largest downward force and the largest upward force that can be exerted by the rod.

Problem 4.115

Solution

Step One: Draw Free Body Diagrams

Free body diagram

Step Two: Centroid of the cross section

Consider the cross section and centroid of the hanger:

Areas of each section,

$A_1=3*0.75=2.25i{n}^2$

(6.5-1)

$A_2=3*1=3i{n}^2$

(6.5-2)

$A_3=3*0.75=2.25i{n}^2$

(6.5-3)

Total area of the cross section

$A=A_1+A_2+A_3=2.25+3+2.25=7.5i{n}^2$

(6.5-4)

Distance of the centroid of all sections from the left edge

${\overline{y}}_1=1+\frac{3}{2}=2.5in$

(6.5-5)

${\overline{y}}_2=\frac{1}{2}=0.5in$

(6.5-6)

${\overline{y}}_3=1+\frac{3}{2}=2.5in$

(6.5-7)

Distance of the centroid of the total cross section from the left edge

$\overline{y}=\frac{A_1{\overline{y}}_1+A_2{\overline{y}}_2+A_3{\overline{y}}_3}{A}=\frac{(2.25i{n}^2)(2.5in)+(3i{n}^2)(0.5in)+(2.25i{n}^2)(2.5in)}{7.5i{n}^2}=1.7in$

(6.5-8)

Step Three: Moment of Inertia

Moment of inertia of a rectangular cross section about a point.

$I=\frac{b{d}^3}{12}+A{d}_1^2$

(6.5-9)

Here, b and d are the breath and width of the rectangular respectively, A is the area of the cross section, and $d_1$ is the distance between the centroid of the cross section and the required point.
Moment of inertia of all section about the centroid of the total cross section,

$I_1=\frac{b_1{d}_1^3}{12}+A_1({\overline{y}}_1-\overline{y}{)}^2=\frac{(0.75in)(3in{)}^3}{12}+(2.25i{n}^2)[(2.5-1.7)in{]}^2=3.1275i{n}^4$

(6.5-10)

$I_2=\frac{b_2{d}_2^3}{12}+A_2({\overline{y}}_2-\overline{y}{)}^2=\frac{(3in)(1in{)}^3}{12}+(3i{n}^2)[(0.5-1.7)in{]}^2=4.57i{n}^4$

(6.5-11)

$I_3=I_1=3.1275i{n}^4$

(6.5-12)

Total moment of inertia of the cross section

$I=I_1+I_2+I_3=(3.1275i{n}^4)+(4.57i{n}^4)+(3.1275i{n}^4)=10.825i{n}^4$

(6.5-13)

Distance of the force from the centroid of the cross section

$d=(4in)+(1.5in)-(1.7in)=3.8in$

(6.5-14)

Let the maximum possible force be P.
Bending moment created due to the force P about the centroid,

$M=-Pd$

(6.5-15)

The negative sign is introduced as the bending moment is in the opposite direction.
The normal stress due to bending at point A,

$({\sigma }_x{)}_{bending}=-\frac{My}{I}$

(6.5-16)

Here, y is the distance of the centroid from the point of consideration.
The normal stress due to centric load,

$({\sigma }_x{)}_{centric}=\frac{P}{A}$

(6.5-17)

Step Four: Maximum acting downward force

Thus, the total normal stress acting at point A, and the force acting is downwards, the normal stress would be

${\sigma }_x=({\sigma }_x{)}_{centric}+({\sigma }_x{)}_{bending}=\frac{P}{A}-\frac{My}{I}$

(6.5-18)

In this case, as the point of consideration moves from the left edge to the right edge the normal stress goes from negative to positive direction, as the magnitude y increases.
Thus, the maximum possible positive stress occurs at the right edge of the cross section.
Here, the distance of the right edge from the centroid, y=+2.3in.
Substitute M=-Pd in equation(6.5-18)

${\sigma }_x=\frac{P}{A}+\frac{Pdy}{I}=P(\frac{1}{A}+\frac{yd}{I})$

(6.5-19)

$P=\frac{{\sigma }_x}{\frac{1}{A}+\frac{yd}{I}}$

(6.5-20)

For the maximum force,

$P_{max}=\frac{{\sigma }_{x,max}}{(\frac{1}{A}+\frac{yd}{I})}=\frac{5ksi}{[\frac{1}{7.5i{n}^2}+\frac{(+2.3in)(3.2in)}{10.825i{n}^4}]}\left|\frac{1kip/i{n}^2}{1ksi}\right|=6.15kips$

(6.5-21)

Similarly, the maximum possible negative stress occurs at the right edge of the cross section.
Here, the distance of the left edge from the centroid, y=-1.7in.

$P_{max}=\frac{-12ksi}{[\frac{1}{7.5i{n}^2}+\frac{(-1.7in)(3.2in)}{10.825i{n}^4}]}\left|\frac{1kip/i{n}^2}{1ksi}\right|=+32.50kips$

(6.5-22)

As the force which has the lower magnitude holds the design condition, the maximum allowable downward force is

$P_{max}=6.15kips$

Step Five: Maximum acting upward force

If the force acting is upwards, the normal stress would be

${\sigma }_x=\frac{-P}{A}-\frac{(-M)y}{I}=-(\frac{P}{A}-\frac{My}{I})$

(6.5-23)

In this case, as the point of consideration moves from the left edge to the right edge the normal stress goes from positive to negative direction, as the magnitude of y increases.
Thus, the maximum possible positive stress occurs at the left edge of the cross section.
Here, the distance of the left edge from the centroid, y =-1.7in.

Substitute M=-Pd in equation(6.5-23)

${\sigma }_x=-(\frac{P}{A}+\frac{Pdy}{I})=-P(\frac{1}{A}+\frac{yd}{I})$

(6.5-24)

$P=\frac{-{\sigma }_x}{\frac{1}{A}+\frac{yd}{I}}$

(6.5-25)

For the maximum force,

$P_{max}=\frac{-{\sigma }_{x,max}}{(\frac{1}{A}+\frac{yd}{I})}=\frac{-(+5ksi)}{[\frac{1}{7.5i{n}^2}+\frac{(-1.7in)(3.2in)}{10.825i{n}^4}]}\left|\frac{1kip/i{n}^2}{1ksi}\right|=13.54kips$

(6.5-26)

Similarly, the maximum possible negative stress occurs at the right edge of the cross section.
Here, the distance of the left edge from the centroid, y=(4in)-(1.7in)=+2.3in

$P=\frac{-(-12ksi)}{[\frac{1}{7.5i{n}^2}+\frac{(+2.3in)(3.2in)}{10.825i{n}^4}]}\left|\frac{1kip/i{n}^2}{1ksi}\right|=14.76kips$

(6.5-27)

As the force which has the lower magnitude holds the design condition, the maximum allowable upward force is

$P_{max}=13.54kips$

References

Beer, F. P., Johnston, E. R., Jr., DeWolf, J. T., & Mazurek, D. F. (2012). Mechanics of materials (6th ed.). New York, NY: McGraw Hill.

External Links

• Online Calculator for Moment of Inertia