# University of Florida/Egm3520/Mom-s13-team4-R5

Report 5

## Problem 5.1

P4.7, Beer 2012

### Problem Statement

Two W4x13 rolled sections are welded together as shown. For the steel alloy used: ${\displaystyle \sigma _{y}=36ksi}$, ${\displaystyle \sigma _{U}=58ksi}$ and a factor of safety of 3.0

### Objective

Determine the largest couple that can be applied when the assembly is bent about the z axis.

### Step 1

Draw dimensions from appendix C.

### Step 2

From appendix C for W4x13:

The area is equal to ${\displaystyle A=3.83in^{2}}$

The moment of inertia about x is equal to ${\displaystyle I_{x}=11.3in^{4}}$

The base is equal to ${\displaystyle b=4.06in}$

### Step 3

The parallel axis theorem gives us the following

${\displaystyle I_{x}=I_{x'}-A(x)^{2}}$

${\displaystyle I_{x'}}$ being the moment about the neutral axis

Solving for the moment of inertia about the neutral axis, we find

${\displaystyle I_{x'}=I_{x}+A(x)^{2}}$

Since there are two sections and ${\displaystyle x={\frac {b}{2}}={\frac {4.16}{2}}=2.08in}$ the moment of inertia of the two sections about the neutral axis is

${\displaystyle I_{s}=2[I_{x}+A*(2.08)^{2}]=2[11.3+3.83*(208)^{2}]=55.7in^{4}}$

### Step 4

Allowable stress is equal to the ultimate stress divided by the factor safety

${\displaystyle \sigma _{a}={\frac {\sigma _{u}}{F.S.}}={\frac {58}{3}}=19.33ksi}$

### Step 5

${\displaystyle M_{max}={\frac {\sigma _{m}*I_{s}}{c}}={\frac {19.33*55.7}{4.16}}=259kip*in}$

The largest couple that can be applied when the assembly is bent about the z axis is 1259 kip*in

### Honor Pledge

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

## Problem 5.2

P4.8, Beer 2012

### Problem Statement

Two W4x13 rolled sections are welded together as shown. For the steel alloy used: ${\displaystyle \sigma _{y}=36ksi}$, ${\displaystyle \sigma _{U}=58ksi}$ and a factor of safety of 3.0

### Objective

Determine the largest couple that can be applied when the assembly is bent about the z axis.

### Step 1

Draw dimensions from appendix C.

### Step 2

From appendix C for W4x13:

The area is equal to ${\displaystyle A=3.83in^{2}}$

The moment of inertia about y is equal to ${\displaystyle I_{Y}=3.86in^{4}}$

The base is equal to ${\displaystyle b=4.06in}$

### Step 3

Allowable stress is equal to the ultimate stress divided by the factor safety

${\displaystyle \sigma _{a}={\frac {\sigma _{u}}{F.S.}}={\frac {58}{3}}=19.33ksi}$

### Step 4

The parallel axis theorem gives us the following

${\displaystyle I_{y}=I_{n}-A(y)^{2}}$

${\displaystyle I_{n}}$ being the moment about the neutral axis

Solving for the moment of inertia about the neutral axis, we find

${\displaystyle I_{n}=I_{y}+A(y)^{2}}$

Since there are two sections and ${\displaystyle y={\frac {b}{2}}={\frac {4.06}{2}}=2.03in}$ the moment of inertia of the two sections about the neutral axis is

${\displaystyle I_{s}=2[I_{y}+A*(2.03)^{2}]=2[3.86+3.83*4.12]=39.29in^{4}}$

### Step 5

The largest distance from from the centroid to either side is ${\displaystyle c=b=4.06in}$

${\displaystyle M_{max}={\frac {\sigma _{a}*I_{s}}{c}}={\frac {19.33*39.29}{4.06}}=187.1kip*in}$

The largest couple that can be applied when the assembly is bent about the z axis is 187.1 kip*in

### Honor Pledge

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

## Problem 5.3

P4.13, Beer 2012

### Problem Statement

A beam of the cross section shown is bent about a horizontal axis and that the bending moment is 6 kN*m.

### Objective

Determine the total force acting on the shaded portion of the web.

### Step 1

To determine the total force acting on the shades area

we need to find the distribution of throught the shades area

the distribution would be:

${\displaystyle dF=\sigma _{x}\times dA}$

${\displaystyle \Rightarrow -{\frac {My}{I}}\times dA}$

${\displaystyle \Rightarrow F=\int -{\frac {My}{I}}\times dA}$

${\displaystyle \Rightarrow F=-{\frac {My}{I}}\int ydA}$

${\displaystyle \Rightarrow F=-{\frac {M}{I}}\times {\bar {y}}\times A}$

### Step 2

We have

${\displaystyle A=0.072\times 0.09=0.00648m^{2}}$

and ${\displaystyle {\bar {y}}=0.045m}$

the centroidal Moment of Inertia is

${\displaystyle I=\sum (I_{x}+Ad^{2})}$

${\displaystyle \Rightarrow I={\frac {1}{12}}\times b_{1}\times h_{1}^{3}+A_{1}\times d_{1}^{2}+{\frac {1}{12}}\times b_{2}\times h_{2}^{3}+A_{2}\times d_{2}^{2}}$

${\displaystyle \Rightarrow I={\frac {1}{12}}\times 216\times 36^{3}+216\times 36\times 36^{2}+{\frac {1}{12}}\times 72\times 90^{3}+90\times 72\times

45^{2}=28.42*10^{6}mm^{4}=28.42*10^{-6}m}$

then ${\displaystyle F={\frac {6000\times 0.00648\times 0.045}{28.42*10^{-6}}}=61.562kN}$

### Honor Pledge

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

## Problem 5.4

P4.16, Beer 2012

### Problem Statement

The beam shown is made of a nylon for which the allowable stress is 24 MPa in tension and 30 MPa in compression.

### Objective

Determine the largest couple M that can be applied to the beam for ${\displaystyle d=40mm}$

### Givens

b = 40mm
s = 15mm
d = 30mm
h = d-s = 15mm
t = 20mm
${\displaystyle {\overline {y_{1}}}}$=?
${\displaystyle {\overline {y_{2}}}}$=?

### Step 1

In order to find the Neutral Axis, we must find the centroid of the T-shape cross-section

${\displaystyle {\overline {y_{1}}}={\frac {(40mm*15mm*7.5mm)+(25mm*20mm*27.5mm)}{40mm*15mm+25mm*20mm}}=16.59mm}$

### Step 2

Now We solve for ${\displaystyle {\overline {y_{2}}}}$

${\displaystyle {\overline {y_{2}}}=40mm-{\overline {y_{1}}}=23.41mm}$

### Step 3

Now we must solve for the Moment of Inertia of the T Shape:
${\displaystyle I={\frac {1}{3}}[t*{\overline {y_{2}}}^{3}+b*{\overline {y_{1}}}^{3}-(b-t)({\overline {y_{1}}}-s)^{3}]}$

${\displaystyle I={\frac {1}{3}}[20mm*23.41^{3}mm+40mm*16.59^{3}mm-(40mm-20mm)(16.59mm-15mm)^{3}]=146,383mm^{4}}$

### Step 4

We can calculate the maximum tensile strength, given that our maximum compression stress is 30Mpa.
${\displaystyle \sigma _{Tmax}={\frac {\overline {y_{1}}}{\overline {y_{2}}}}*30Mpa={\frac {16.59}{23.41}}*30Mpa=21.26Mpa}$

### Step 5

Since ${\displaystyle \sigma _{Tmax}<24Mpa}$, the maximum stress is seen through compression. Therefore, we will use that compression stress in the elastic flexural formula:

${\displaystyle {\frac {M}{I}}={\frac {\sigma _{C}}{\overline {y_{2}}}}}$

### Step 6

Therefore, the largest couple moment that can be applied to the beam is as follows:

${\displaystyle M={\frac {146,383*10^{-12}m^{4}*30*10^{6}Pa}{.02341m}}=187.59N*m}$

### Honor Pledge

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

## Problem 5.5

P4.20, Beer 2012

### Problem Statement

The extruded beam shown has allowable stress is 120 MPa in tension and 150 MPa in compression.

### Objective

Determine the largest couple M that can be applied.

### Solution

#### Step 1

The centroid of a trapezoid is given by

${\displaystyle y={\frac {b+2a}{3(a+b)}}h}$
where a = 80 mm, b = 40 mm, and h = 54 mm

so

${\displaystyle y={\frac {40mm+2*80mm}{3(80mm+40mm)}}54mm=30mm}$

#### Step 2

Splitting the trapezoid into 2 triangles and a rectangle we can find the Moment of inertia of the trapezoid by

summing the individual moments of inertia.

#### Step 3

The moment of inertia of the triangle is given by:

${\displaystyle I={\frac {1}{36}}bh^{3}={\frac {1}{36}}20mm*54mm^{3}=827089.4mm^{4}}$

The Area of the triangle is:

${\displaystyle A={\frac {1}{2}}bh={\frac {1}{2}}20mm*54mm=540mm^{2}}$

centroid of a triangle is : y = ${\displaystyle {\frac {1}{3}}h={\frac {1}{3}}54mm=18mm}$

so dy = 36mm - 30mm = 6mm

#### Step 4

The moment of inertia of the rectangle is given by:

${\displaystyle I={\frac {1}{12}}bh^{3}={\frac {1}{12}}40mm*54mm^{3}=524880mm^{4}}$

The Area of the rectangle is:

${\displaystyle A=bh=40mm*54mm=2160mm^{2}}$

the centroid of a rectangle is the center so y = 27 mm

so dy = 30 mm - 27 mm = 3 mm

#### Step 5

${\displaystyle I_{xx}=2(I_{triangle}+Ady^{2})+(I_{rectangle}+Ady^{2})}$

${\displaystyle I_{xx}=2(82709.4+540*6^{2})+(82709.4+2160*3^{2})=748619mm^{4}}$

#### Step 6

Applying the Elastic fexural formula to get:

${\displaystyle M={\frac {\sigma }{c}}*I_{xx}}$

Looking at the bottom half of the beam gives us c = 30 mm and ${\displaystyle \sigma =150MPa}$

${\displaystyle M={\frac {150*10^{6}}{30*10^{-3}}}*748619*10^{-12}=3748.1N*m}$

Looking at the top half of the beam gives us c = 24 mm and ${\displaystyle \sigma =120MPa}$

${\displaystyle M={\frac {120*10^{6}}{24*10^{-3}}}*748619*10^{-12}=3743.1N*m}$

The larges couple M is felt by the bottom half

${\displaystyle M_{max}=3748.1N*m}$

### Honor Pledge

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

## Problem 5.6

P3.53, Beer 2012

### Problem Statement

The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is ${\displaystyle 3.7*10^{6}psi}$ for aluminum and ${\displaystyle 5.6*10^{6}psi}$ for brass.

### Objective

Determine the maximum shearing stress (a) in cylinder AB, (b) in cylinder BC.

### Solution

FBD

We need to split the solid shaft AC into two free body diagrams, shaft AB and shaft BC
Given

#### Step 1

In order to find the max shearing stress, we need to determine the Torques at point A and C
${\displaystyle \sum M_{x}=0}$
${\displaystyle \Rightarrow T_{AB}+T_{BC}-T=0\Rightarrow T_{AB}+T_{BC}=12.5kips\cdot in}$

${\displaystyle T_{BC}=3.1189T_{AB}}$

${\displaystyle T_{AB}+3.1889T_{AB}=12.5*10^{3}}$

${\displaystyle T_{BC}=9.51596*10^{3}lb*in}$

${\displaystyle T_{AB}=2.98403*10^{3}lb*in}$

#### Step 2

Find the moment of inertia in each cylinder

${\displaystyle J_{AB}={\frac {\pi }{32}}D^{4}={\frac {\pi }{32}}1.5in^{4}=.497in^{4}}$

${\displaystyle J_{BC}={\frac {\pi }{32}}D^{4}={\frac {\pi }{32}}2in^{4}=1.57in^{4}}$

#### Step 3

Find the max sheer stress in each cylinder

${\displaystyle \tau _{AB}={\frac {T_{AB}C_{AB}}{J_{AB}}}}$

${\displaystyle \tau _{AB}={\frac {T_{AB}C_{AB}}{J_{AB}}}={\frac {2.98403kip*in*.75in}{.497in^{4}}}=.281509kip*in}$

${\displaystyle \tau _{AC}={\frac {T_{AC}C_{AC}}{J_{AC}}}={\frac {9.51596kip*in*1in}{25.13in^{4}}}=.378669kip*in}$

### Honor Pledge

On our honor, we did not do this problem on our own, without looking at the solutions in previous semesters or other online solutions.