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

Problem 5.1
P4.7, Beer 2012

Problem Statement
Two W4x13 rolled sections are welded together as shown. For the steel alloy used: $$\sigma _y=36 ksi$$, $$\sigma _U= 58 ksi$$ 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 $$A=3.83in^2$$ The moment of inertia about x is equal to $$I_x=11.3 in^4$$ The base is equal to $$b=4.06 in$$

Step 3
The parallel axis theorem gives us the following $$I_x=I_{x'}-A(x)^2$$ $$I_{x'}$$ being the moment about the neutral axis Solving for the moment of inertia about the neutral axis, we find $$I_{x'}=I_x+A(x)^2$$ Since there are two sections and $$x=\frac{b}{2}=\frac{4.16}{2}=2.08 in$$ the moment of inertia of the two sections about the neutral axis is $$I_s=2[I_{x}+A*(2.08)^2]=2[11.3+3.83*(208)^2]=55.7 in^4$$

Step 4
Allowable stress is equal to the ultimate stress divided by the factor safety

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

Step 5
$$M_{max}=\frac{\sigma _m*I_s}{c}=\frac{19.33*55.7}{4.16}=259 kip*in$$ 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: $$\sigma _y=36 ksi$$, $$\sigma _U= 58 ksi$$ 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 $$A=3.83in^2$$ The moment of inertia about y is equal to $$I_Y=3.86 in^4$$ The base is equal to $$b=4.06 in$$

Step 3
Allowable stress is equal to the ultimate stress divided by the factor safety

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

Step 4
The parallel axis theorem gives us the following $$I_y=I_n-A(y)^2$$ $$I_n$$ being the moment about the neutral axis Solving for the moment of inertia about the neutral axis, we find $$I_n=I_y+A(y)^2$$ Since there are two sections and $$y=\frac{b}{2}=\frac{4.06}{2}=2.03 in$$ the moment of inertia of the two sections about the neutral axis is $$I_s=2[I_y+A*(2.03)^2]=2[3.86+3.83*4.12]=39.29 in^4$$

Step 5
The largest distance from from the centroid to either side is $$c=b=4.06 in$$ $$M_{max}=\frac{\sigma _a*I_s}{c}=\frac{19.33*39.29}{4.06}=187.1 kip*in$$ 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:

$$dF=\sigma _x\times dA$$

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

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

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

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

Step 2
We have

$$A=0.072\times 0.09=0.00648m^2$$

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

the centroidal Moment of Inertia is

$$I=\sum (I_x+Ad^2)$$

$$\Rightarrow I=\frac{1}{12}\times b_1\times h_1^3+A_1\times d_1^2+\frac{1}{12}\times b_2\times h_2^3+A_2\times d_2^2$$

$$\Rightarrow I=\frac{1}{12}\times 216\times 36^3+216\times 36\times 36^2+\frac{1}{12}\times 72\times 90^3+90\times 72\times 45^2=28.42*10^6mm^4=28.42*10^{-6}m$$

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

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 $$d=40 mm$$

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

Step 1
In order to find the Neutral Axis, we must find the centroid of the T-shape cross-section $$ \overline{y_{1}} = \frac{(40mm*15mm*7.5mm)+(25mm*20mm*27.5mm)}{40mm*15mm+25mm*20mm} = 16.59mm $$

Step 2
Now We solve for $$\overline{y_{2}} $$ $$\overline{y_{2}} = 40mm - \overline{y_{1}} = 23.41mm $$

Step 3
Now we must solve for the Moment of Inertia of the T Shape: $$ I = \frac{1}{3}[t*\overline{y_{2}}^{3}+ b*\overline{y_{1}}^{3}-(b-t)(\overline{y_{1}}-s)^{3}] $$ $$ I = \frac{1}{3}[20mm*23.41^{3}mm+ 40mm*16.59^{3}mm-(40mm-20mm)(16.59mm-15mm)^{3}] = 146,383mm^{4} $$

Step 4
We can calculate the maximum tensile strength, given that our maximum compression stress is 30Mpa. $$ \sigma_{Tmax} = \frac{\overline{y_{1}}}{\overline{y_{2}}}*30Mpa = \frac{16.59}{23.41}*30Mpa = 21.26 Mpa $$

Step 5
Since $$ \sigma_{Tmax} < 24 Mpa $$, the maximum stress is seen through compression. Therefore, we will use that compression stress in the elastic flexural formula: $$ \frac{M}{I}= \frac{\sigma_{C}}{\overline{y_{2}}}$$

Step 6
Therefore, the largest couple moment that can be applied to the beam is as follows: $$ M = \frac{146,383*10^{-12}m^{4}*30*10^{6}Pa}{.02341m} = 187.59 N*m$$

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.

Step 1
The centroid of a trapezoid is given by $$ y = \frac{b+2a}{3(a+b)}h $$ where a = 80 mm, b = 40 mm, and h = 54 mm

so

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

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:

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

The Area of the triangle is:

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

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

so dy = 36mm - 30mm = 6mm

Step 4
The moment of inertia of the rectangle is given by:

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

The Area of the rectangle is:

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

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

so dy = 30 mm - 27 mm = 3 mm

Step 5
$$ I_{xx} = 2 (I_{triangle} + Ady^{2}) + (I_{rectangle} + Ady^{2}) $$

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

Step 6
Applying the Elastic fexural formula to get:

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

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

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

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

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

The larges couple M is felt by the bottom half

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

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 $$ 3.7*10^{6} psi$$ for aluminum and $$5.6*10^{6} psi$$ 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 $$\sum M_x=0$$ $$\Rightarrow T_{AB}+T_{BC}-T=0 \Rightarrow T_{AB}+T_{BC}=12.5 kips\cdot in$$

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

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

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

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

Step 2
Find the moment of inertia in each cylinder

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

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

Step 3
Find the max sheer stress in each cylinder

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

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

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

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.