# Mechanics of materials/Problem set 5

## Problem 5.1 (Problem 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

Two W4 x 13 rolled sections are welded together. Determine that largest couple that can be applied about the z-axis when the assembly is bent.
The steel alloy used has properties ${\displaystyle \sigma _{\gamma }=36ksi}$  and ${\displaystyle \sigma _{U}=58ksi}$ . Use a safety factor of 3.0.

Obtained from Appendix B in Beer, 2012 are the properties of W4x13 Rolled-steel:
Area, ${\displaystyle A}$ = 3.83 in2
Width, ${\displaystyle W}$ = 4.16 in
Moment of Inertia (with respect of y-axis), ${\displaystyle I_{y}}$ = 11.3 in4

### Solution

#### Step One: Divide assembly in two

The first step towards an easy and straightforward solution is the division of the given assembly at the midplane of the y-axis, parallel to the z-axis. This eases the mathematics involved in the solution, taking advantage of the assembly's inherent symmetry. The division is shown below, where the new plane is called the A-A axis.

#### Step Two: Solve for moment of inertia about A-A

The moment of inertia about A-A is given by Equation 5.1-1, derived from the parallel-axis theorem.

 ${\displaystyle \displaystyle I_{A-A}=I_{x}+Ad^{2}}$ (5.1-1)

where, Ix is the moment of inertia through the centroid with respect to the x-axis and d2 is the square of the distance from the centroid of the shape to the A-A axis. For this particular shape, we can calculate the coordinates of its centroid from symmetry.

Substituting in the values for Ix, A, and d2 into Equation(5.2-1), we obtain

${\displaystyle I_{a-a}=11.3in^{4}+(3.83in^{2})(2.08in)^{2}}$
${\displaystyle I_{a-a}=27.9in^{4}}$

#### Step Three: Extrapolate simplified solution to the entire assembly

The total moment of inertia of both sections together is the sum of the individual moments with respect to the same axis. Thus, for the original shape, the total moment of inertia is

${\displaystyle I_{z}=2I_{A-A}=55.8in^{4}}$

because both sections are identical and symmetrical.

The bending moment, or couple, is given by

 ${\displaystyle \displaystyle M={\frac {I\sigma _{all}}{c}}}$ (5.1-2)

for which ${\displaystyle \sigma _{all}}$  is the maximum allowable stress which is calculated from ${\displaystyle \sigma _{U}}$  and the factor of safety using the following equation

 ${\displaystyle \displaystyle \sigma _{all}={\frac {\sigma _{U}}{F.S}}={\frac {58ksi}{3.0}}=19.33ksi}$ (5.1-3)

Returning to Eq. 5.1-2 and substituting in the values for ${\displaystyle I,\sigma _{all},}$  and ${\displaystyle c}$ , which is the width, yields

${\displaystyle M={\frac {I\sigma _{all}}{c}}}$


${\displaystyle M={\frac {57.8ksi\times 19.33ksi}{4.16in}}}$

 ${\displaystyle \displaystyle M=259kip\cdot in}$

## Problem 5.2 (Problem 4.8 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

Two W4 x 13 rolled sections are welded together. Determine that largest couple that can be applied about the z-axis when the assembly is bent.
The steel alloy used has properties ${\displaystyle \sigma _{\gamma }=36ksi}$  and ${\displaystyle \sigma _{U}=58ksi}$ . Use a safety factor of 3.0.

Obtained from Appendix B in Beer, 2012 are the properties of W4x13 Rolled-steel:
Area, ${\displaystyle A}$ = 3.83 in2
Width, ${\displaystyle W}$ = 4.060 in
Moment of Inertia (with respect of y-axis), ${\displaystyle I_{y}}$ = 3.86 in4

### Solution

The first step in the solution is to realize that this is a symmetrical composite shape, and the shape can be split in two to facilitate the solution process.
It is favorable to divide the shape by some axis that will conveniently make the problem easier to solve. In this case, it will be in the middle, across the z-axis.
The axis A-A across the bottom of the top half now corresponds to this slice along the z-axis.

Here, a new point of reference is assigned.

The next step is to calculate the moment of inertia about this new axis, A-A. This is done using the parallel-axis theorem, given by:

 ${\displaystyle \displaystyle I_{A-A}=I_{y}+Ad^{2}}$ (5.2-1)

where, Iy is the moment of inertia through the centroid of the shape with respect to the y-axis and d2 is the square of the distance from the centroid of the shape to the axis for which the moment of inertia is being taken. For this particular shape, we can calculate the coordinates of its centroid from symmetry.

Substituting in the values for Iy, A, and d2 into Equation(5.2-1), we obtain

${\displaystyle I_{a-a}=3.86in^{4}+(3.83in^{2})(2.03in)^{2}}$
${\displaystyle I_{a-a}=19.643in^{4}}$

Now, the total moment of inertia of both sections together is the sum of the individual moments with respect to the same axis. Thus, for the original shape, the total moment of inertia is

${\displaystyle I=2I_{a-a}=39.286in^{4}}$

because both sections are identical and symmetrical.

The bending moment, or couple, is given by

 ${\displaystyle \displaystyle M={\frac {I\sigma _{all}}{c}}}$ (5.2-2)

for which ${\displaystyle \sigma _{all}}$  is the maximum allowable stress which is calculated from ${\displaystyle \sigma _{U}}$  and the factor of safety using the following equation

 ${\displaystyle \displaystyle \sigma _{all}={\frac {\sigma _{U}}{F.S}}={\frac {58ksi}{3.0}}=19.33ksi}$ (5.2-3)

Returning to Eq. 5.2-2 and substituting in the values for ${\displaystyle I,\sigma _{all},}$  and ${\displaystyle c}$ , which is the width, yields

${\displaystyle M={\frac {I\sigma _{all}}{c}}}$


${\displaystyle M={\frac {39.286ksi\times 19.33ksi}{4.06in}}}$

 ${\displaystyle \displaystyle M=187.1kip\cdot in}$

## Problem 5.3 (Problem 4.13 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

Knowing that the beam of the cross section below is bent about the horizontal axis with a bending moment, the total force acting on the shaded portion of the T-beam's web is to be determined.

#### Given

Area of T-beam's flange (Area 1): ${\displaystyle 216mm\times 36mm}$

Area of T-beam's web (Area 2): ${\displaystyle 72mm\times 108mm}$

Bending moment about horizontal axis: ${\displaystyle M_{z}=6kN\cdot m}$

### Solution

#### Derivation of Force on Shaded Area of T-beam's Web

In order to find the force on the shaded area of the T-beam's web due to the bending moment, the stress distribution must be used, which is given by the bending stress formula:

 ${\displaystyle \displaystyle \sigma _{x}=-{\frac {M_{z}y}{I_{z}}}}$ (5.3-1)

By using a small area element ${\displaystyle \mathrm {d} A}$ , the force on the shaded region can be calculated by:

 ${\displaystyle \displaystyle \mathrm {d} F=\sigma _{x}\mathrm {d} A=-{\frac {M_{z}y}{I_{z}}}\,\mathrm {d} A}$ (5.3-2)

Which, when integrated, solves for the force we're looking for:

 ${\displaystyle \displaystyle F=\int \!\mathrm {d} F=-\int \!{\frac {M_{z}y}{I_{z}}}\,\mathrm {d} A=-{\frac {M_{z}}{I_{z}}}\int \!y\,\mathrm {d} A=-{\frac {M_{z}}{I_{z}}}{\bar {y}}^{*}A^{*}}$ (5.3-3)

where ${\displaystyle {\bar {y}}^{*}}$  and ${\displaystyle A^{*}}$  are the distance of the centroid from the neutral axis and area of the shaded region, respectively.

#### Location of Centroids and Distance From Neutral Axis for the Flange and Web

The T-beam is split up into two distinct rectangles which comprise the area of the flange (top rectangle) and the area of the web (bottom rectangle). The distance of these rectangle's centroids from the neutral axis are therefore:

${\displaystyle d_{1}=54mm-{\frac {36mm}{2}}=36mm}$

${\displaystyle d_{2}=108mm-(54mm-36mm)-{\frac {108mm}{2}}=36mm}$

#### Calculation of 2nd Moment of Inertia for Beam Cross Section

The 2nd moment of inertia for the entire beam cross section can be calculated using the parallel axis theorem,

 ${\displaystyle \displaystyle I_{z}=\sum (I+Ad^{2})}$ (5.3-4)

where ${\displaystyle I}$  for a rectangle is:

 ${\displaystyle \displaystyle I={\frac {bh^{3}}{12}}}$ (5.3-5)

Using (5.3-4) and (5.3-5), the 2nd moment of inertia for the T-beam's flange (top rectangle) can be written as:

 ${\displaystyle \displaystyle I_{1}={\frac {1}{12}}b_{1}h_{1}^{3}+A_{1}d_{1}^{2}}$ (5.3-6)

which gives:

${\displaystyle I_{1}={\frac {1}{12}}(216)(36)^{3}+(216)(36)(36)^{2}=10.92\times 10^{6}mm^{4}}$

And the 2nd moment of inertia for the T-beam's web (bottom rectangle) can be written as:

 ${\displaystyle \displaystyle I_{2}={\frac {1}{12}}b_{2}h_{2}^{3}+A_{2}d_{2}^{2}}$ (5.3-7)

which gives:

${\displaystyle I_{2}={\frac {1}{12}}(72)(108)^{3}+(72)(1108)(36)^{2}=17.64\times 10^{6}mm^{4}}$

Therefore the total 2nd moment of inertia is the combination of both the flange and web moments:

 ${\displaystyle \displaystyle I_{z}=I_{1}+I_{2}}$ (5.3-8)

which gives:

${\displaystyle I_{z}=10.92\times 10^{6}mm^{4}+17.64\times 10^{6}mm^{4}=28.56\times 10^{-6}m^{4}}$

#### Location of Centroid and Area of the Web's Shaded Region

The shaded region of the T-beam's web has a width of:

${\displaystyle 72mm}$

and a height of:

${\displaystyle 108mm-(54mm-36mm)=90mm}$

Which means the shaded region has an area of:

${\displaystyle A^{*}=72mm\times 90mm=6480mm^{2}}$

The distance of the centroid from the neutral axis is:

${\displaystyle {\bar {y}}^{*}=90mm-{\frac {90mm}{2}}=45mm}$

Therefore, the product of the shaded region's area and centroid distance from the neutral axis is:

${\displaystyle {\bar {y}}^{*}A^{*}=45mm\times 6480mm^{2}=291.6\times 10^{-6}m}$

#### Calculation of Force on Shaded Area of T-beam's Web

The force on the shaded region of the T-beam's web can now be calculated using equation (5.3-3) by substituting all relevant values:

 ${\displaystyle \displaystyle F=-{\frac {M_{z}}{I_{z}}}{\bar {y}}^{*}A^{*}}$ (5.3-3)

which gives:

${\displaystyle F={\frac {(6\times 10^{3})(291.6\times 10^{-6})}{28.56\times 10^{-6}}}=61.3\times 10^{3}N}$

 ${\displaystyle \displaystyle F=61.3kN}$

## Problem 5.4 (Problem 4.16 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 maximum allowable stress for the beam pictured is 24MPa in tension and 30MPa in compression.

For d=40mm, find the largest couple M which can be applied to the beam.

#### Given

Maximum stress in compression: ${\displaystyle \sigma _{max,C}=30\,MPa}$
Maximum stress in tension: ${\displaystyle \sigma _{max,T}=24\,MPa}$
${\displaystyle d=40\,mm}$

### Solution

#### Analysis

Free body diagram of the beam:

#### Centroid

Splitting the bar's cross section into two rectangles, 1 and 2, with 1 being the upper section and 2 being the lower section

${\displaystyle \displaystyle A_{1}=15\,mm\times 40\,mm=600\,mm^{2}}$
${\displaystyle A_{2}=20\,mm\times 25\,mm=500\,mm^{2}}$
${\displaystyle {\bar {y}}_{1}=25+{\frac {15}{2}}=32.5\,mm}$
${\displaystyle {\bar {y}}_{2}=25/2=12.5\,mm}$

We can use these to find the neutral axis (centroid) for the entire cross section with this relationship

 ${\displaystyle \displaystyle {\bar {Y}}\sum A=\sum {\bar {y}}A}$ (5.4-1)

 ${\displaystyle \displaystyle {\bar {Y}}=23.409\,mm}$ (5.4-2)

#### Centroidal Moment of Inertia

Using the parallel axis theorem, the centroidal moment of inertia can be represented as

 ${\displaystyle \displaystyle I=\sum (I+Ad^{2})}$ (5.4-3)

where ${\displaystyle I}$  for a rectangle is

 ${\displaystyle \displaystyle I={\frac {bh^{3}}{12}}}$ (5.4-4)

Now, inserting the values for the entire bar

 ${\displaystyle \displaystyle I=[{\frac {(40)(15^{3})}{12}}+(600)(32.5-23.409)^{2}]+[{\frac {(20)(25^{3})}{12}}+(500)(12.5-23.409)^{2}]}$ (5.4-5)

 ${\displaystyle \displaystyle I=146.4\times 10^{-9}\,m^{4}}$ (5.4-6)

#### Stress

Because of the direction the bar is being bent due to the applied moment, the top of the beam will undergo compression and the bottom of the bar will undergo tension. So, when calculating the maximum stress due to compression, the distance from the neutral axis to the top of the bar will be used. Likewise, the distance from the neutral axis to the bottom of the bar will be used in the calculation for maximum stress due to tension.

 ${\displaystyle \displaystyle \sigma _{max,C}={\frac {Mc}{I}}\Rightarrow M={\frac {\sigma _{max,C}I}{c}}}$ (5.4-7)

This gives us ${\displaystyle M=187.6\,N\cdot m}$
For tension:

 ${\displaystyle \displaystyle \sigma _{max,T}={\frac {Mc}{I}}\Rightarrow M={\frac {\sigma _{max,T}I}{c}}}$ (5.4-8)

Which gives us ${\displaystyle M=211.8\,N\cdot m}$
The smaller of these two values is our answer.

 ${\displaystyle \displaystyle M=187.6\,N\cdot m}$

(5.4-9)

## Problem 5.5 (Problem 4.20 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

Knowing that for the extruded beam shown the allowable stress is 120 MPa in tension and 150 MPa in compression, determine the largest couple M that can be applied.

#### Given

Stress in tension = 120 MPa
Stress in compression = 150 MPa

### Solution

For a trapezium Centroidal distance from AD is given by

 ${\displaystyle \displaystyle C={\frac {80+2*40}{80+40}}*{\frac {54}{3}}}$ (5.5-1)

 ${\displaystyle \displaystyle C=24mm}$ (5.5-2)

 ${\displaystyle \displaystyle A={\frac {1}{2}}(80+40)*54}$ (5.5-3)

 ${\displaystyle \displaystyle A=3240mm^{2}}$ (5.5-4)

 ${\displaystyle \displaystyle I_{BCEF}={\frac {1}{3}}*40*54^{3}}$ (5.5-5)

 ${\displaystyle \displaystyle I_{BCEF}=2099520mm^{4}}$ (5.5-6)

Moment of Inertia of triangle ABF about AB (AB=CD)

 ${\displaystyle \displaystyle I_{ABF}={\frac {1}{12}}*20*54^{3}}$ (5.5-7)

 ${\displaystyle \displaystyle I_{ABF}=262440mm^{4}}$ (5.5-8)

Base of each Triangle

 ${\displaystyle \displaystyle Base={\frac {80-40}{2}}}$ (5.5-9)

 ${\displaystyle \displaystyle Base=20mm}$ (5.5-9)

Moment of Inertia of triangle CDE about CD

 ${\displaystyle \displaystyle I_{CDE}={\frac {1}{12}}*20*54^{3}}$ (5.5-10)

 ${\displaystyle \displaystyle I_{CDE}=262440mm^{4}}$ (5.5-11)

 ${\displaystyle \displaystyle I_{AD}=I_{BCEF}+I_{ABF}+I_{ABF}}$ (5.5-12)

 ${\displaystyle \displaystyle I_{AD}=2099520+262440+262440}$ (5.5-13)

 ${\displaystyle \displaystyle I_{AD}=2624400mm^{4}}$ (5.5-14)

Now applying parallel axes theorem of moment of Inertia

 ${\displaystyle \displaystyle I_{AD}=I_{xx}+A{\bar {y}}^{2}}$ (5.5-15)

Now rearrange the equation to give

 ${\displaystyle \displaystyle I_{xx}=I_{AD}-A{\bar {y}}^{2}}$ (5.5-16)

 ${\displaystyle \displaystyle I_{xx}=2624400-3240*24^{2}}$ (5.5-17)

 ${\displaystyle \displaystyle I_{xx}=758160mm^{4}}$ (5.5-18)

Now,

 ${\displaystyle \displaystyle M={\frac {\sigma }{C}}*I_{xx}}$ (5.5-19)

 ${\displaystyle \displaystyle M={\frac {120*10^{6}}{24810^{-}3}}*{\frac {758160}{10^{1}2}}}$ (5.5-20)

 ${\displaystyle \displaystyle M=37908*{\frac {10^{8}}{10^{9}}}}$ (5.5-21)

 ${\displaystyle \displaystyle M=3790.8N*m}$ (5.5-22)

Bottom surface is in compression, for allowable compression stress of 150 MPa
Here

 ${\displaystyle \displaystyle C=54-24}$ (5.5-23)

 ${\displaystyle \displaystyle C=30mm}$ (5.5-24)

Couple M will be

 ${\displaystyle \displaystyle M={\frac {\sigma }{C}}*I_{xx}}$ (5.5-25)

 ${\displaystyle \displaystyle M={\frac {150*10^{6}}{30*10^{-}3}}*{\frac {758160}{10^{1}2}}}$ (5.5-26)

 ${\displaystyle \displaystyle M=37908*{\frac {10^{8}}{10^{9}}}}$ (5.5-27)

 ${\displaystyle \displaystyle M=3790.8N*m}$ (5.5-28)

Both the values given in equations (5.5-22) and (5.5-28) Therefore

 ${\displaystyle \displaystyle M=3790.8N*m}$

(5.5-29)

## Problem 5.6 (Problem 3.53 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 solid cylinders AB and BC are bonded together and are attached to fixed supports at A and C. Know the modulus of both aluminum and brass, determine the maximum shearing stress in both AB and BC.

### Given

Modulus for aluminum is 3.7*10^6 psi

Modulus for brass is 5.6*10^6 psi

### Solution

Equilibrium Solution

 ${\displaystyle T_{A}+T_{C}=12.5kip*in}$ (3.53-1)

The compatibility condition

 ${\displaystyle \phi _{B/C}+\phi _{C/B}=0}$ (5.6-2)

 ${\displaystyle ({\frac {TL}{GJ}})_{AB}+({\frac {TL}{GJ}})_{BC}=0}$ (5.6-3)

The polar moment of each section

 ${\displaystyle J_{AB}={\frac {\pi }{2}}L_{AB}^{4}={\frac {\pi }{2}}(1.5in)^{4}=7.95in^{4}}$ (5.6-4)

 ${\displaystyle J_{BC}={\frac {\pi }{2}}L_{BC}^{4}={\frac {\pi }{2}}(2in)^{4}=25.13in^{4}}$ (5.6-5)

The internal torques

 ${\displaystyle T_{AB}=T_{A}}$ (5.6-6)

 ${\displaystyle T_{BC}=-T_{C}}$ (5.6-7)

Substituting into the compatibility condition

 ${\displaystyle T_{A}({\frac {12}{3.7*10^{6}*7.95}})-T_{C}({\frac {18}{5.6*10^{6}*25.13}})}$ (5.6-8)

 ${\displaystyle T_{A}(4.08*10^{-}7)-T_{C}(1.27*10^{-}7)=0}$ (5.6-9)

 ${\displaystyle T_{A}=T_{C}(.3135)}$ (5.6-10)

 ${\displaystyle T_{A}+T_{C}=12.5kip*in}$ (5.6-11)

 ${\displaystyle T_{C}(.3135)+T_{C}=12.5kip*in}$ (5.6-12)

 ${\displaystyle T_{C}=9.517kip*in}$ (5.6-13)

 ${\displaystyle T_{A}=2.983kip*in}$ (5.6-14)

The max sheer stress in cylinder AB

 ${\displaystyle (\tau _{AB})={\frac {T_{A}*c}{J_{AB}}}={\frac {2.98*{\frac {1.5}{2}}}{7.95}}=.2811lb*in}$ (5.6-15)

The max sheer stress in cylinder BC

 ${\displaystyle (\tau _{AC})={\frac {T_{C}*c}{J_{CB}}}={\frac {9.571*{\frac {2}{2}}}{25.13}}=.3787lb*in}$ (5.6-16)

 ${\displaystyle \displaystyle (\tau _{AB})=0.2811lb*in}$

(5.6-17)

 ${\displaystyle \displaystyle (\tau _{AC})=0.3787lb*in}$

(5.6-18)

## 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.