User:Egm3520.s13.team4.diaz/Mom-s13-team4-R6

Problem 6.1
P4.101, Beer 2012

Problem Statement
Knowing that the magnitude of the horizontal force P is 8 kN, determine the stress at (a) point A, (b) point B.

Step 1
With the given quantities: P = 8 kN bend height h = 45 mm cross section width = 30 mm cross section height = 24 mm

we can find the following quanities:

$$ A = h_{cs} * w _{cs} = .03 m * .024 m = 7.2 *10^{-4} m^{2} $$

$$ I = \frac{1}{12}bd^{3} = \frac{1}{12}.03m*.024m^{3} = 3.456 *10^{-8}m^{4}$$

$$ y_{bar} = \frac{h}{2} = .012 m $$

Step 2
The distance from the centroid to the applied force (d) is:

$$ d = h - y_{bar} = .045 m - .012m = .033 m $$

And now we can use the formula to find the Moment

$$ M = Pd = .033 m * 8000 N = 264 N*m $$

Step 3
$$\sigma _{o}= \frac{P}{A} = \frac{8000 N}{7.2 *10^{-4}m^{2}} = 11.11 M Pa $$

since $$ \sigma_{o} $$ is in compression then $$ \sigma_{o} = -11.11 MPa $$

$$\sigma _{bend}= \frac{Mc}{I} = \frac{264 N*m * .012 m}{3.456 *10^{-8}m^{4}} = 91.6 M Pa$$

Step 4
Maxim tensile and compressible stresses:

compression at A -> $$ \sigma _{c} = \sigma _{o} - \sigma _{bend} = -11.11 MPa - 91.6 MPa = -102.7 MPa $$

tension at B -> $$ \sigma _{c} = \sigma _{o} + \sigma _{bend} = -11.11 MPa +91.6 MPa = 80.49 MPa $$

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 6.2
P4.103, Beer 2012

Problem Statement
The vertical portion of the press shown consists of a rectangular tube of wall thickness t= 8 mm. The press has been tightened on wooden planks being glued together until P= 20 kN.



Objective
Determine the stress at (a) point A, (b) point B.

Step 1
The Area

$$A=A_{outer}-A_{inner}=(60*80)*10^{-6}-(44*64)*10^{-6}=1.984*10^{-6}m^2$$

Step 2
The moment of Inertia

$$I=I_{outer}-I_{inner}=\left (\frac{bh^3}{12}\right )_{outer}-\left (\frac{bh^3}{12}\right )_{inner}=\frac{60*80^3}{12}-\frac{44*64^3}{12}=1.599*10^{-6}m^4$$

Step 3
The Centroid



Step 4
The Moment

$$M=P\times d=20000*240*10^{-3}=4800Nm$$

Step 5
a)The stress at A

$$\sigma _A=\sigma_{A_{centric}}+\sigma_{A_{bending}}=\frac{P}{A}+\frac{My}{I}=\frac{20000N}{1.98*10^{-3}m^2}+\frac{4800\ast 0.04}{1.599*10^{-6}}=1.31*10^{10}Pa$$

Step 6
b)The stress at B

$$\sigma _B=\sigma_{B_{centric}}+\sigma_{B_{bending}}=\frac{P}{A}+\frac{My}{I}=\frac{20000N}{1.98*10^{-3}m^2}-\frac{4800*0.04}{1.599*10^{-6}}=-110MPa$$

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 6.3
P4.112, Beer 2012

Problem Statement
An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. The maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight.



Objective
Determine the largest offset that can be used.

Diagrams:
Dimensions:



$$d=0.75in$$ $$t=0.08in$$ $$h= unknown$$

Free Body Diagram:





Stress when the tube is straight $$\sigma _s$$
The inner diameter:

$$d_i=d-2t=0.75in-2*0.08in=0.59in$$

The cross sectional area of the tube:

$$A=\frac{\pi }{4}(d^2-d_i^2)=\frac{\pi}{4}((0.75in)^2-(0.59in)^2)=0.168in^2$$

Stress:

$$\sigma _s=\frac{P}{A}=\frac{P}{0.168in^2}$$

Stress when the tube is bent $$\sigma _b$$
Outer radius c:

$$c=\frac{d}{2}=\frac{0.75in}{2}=0.375in$$

Moment of Inertia I:

$$I=\frac{\pi}{64}(d^4-d_i^4)=\frac{\pi}{64}((0.75in)^4-(0.59in)^4)=9.58*10^{-3}in^4$$

Couple Bending Moment M:

$$M=P\times h=Phsin(90)=Ph$$

Stress:

$$\sigma _b=\frac{P}{A}+\frac{Mc}{I}=\frac{P}{0.168in^2}+\frac{Phc}{9.58*10^{-3}in^4}=\frac{P}{0.168in^2}+\frac{0.375Ph}{9.58*10^{-3}in^4}$$

Maximum offset h
The bent stress must not exceed 4 times the straight stress of the tube:

$$\sigma_b=4\sigma_s$$

$$\frac{P}{0.168in^2}+\frac{0.375Ph}{9.58*10^{-3}in^4}=4*\frac{P}{0.168in^2}$$

$$\frac{1}{0.168in^2}+\frac{0.375h}{9.58*10^{-3}in^4}=4*\frac{1}{0.168in^2}$$

$$h=\frac{9.58*10^{-3}in^4}{0.375}\left (-\frac{1}{0.168in^2}+4*\frac{1}{0.168in^2} \right )$$

$$h=0.456in$$

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 6.4
P4.114, Beer 2012

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



Solution
Step 1

Find the centroid of the cross section



$$\bar{y}=\frac{A_1\bar{y}_1+A_2\bar{y}_2+A_3\bar{y}_3}{A_1+A_2+A_3}$$

$$\bar{y}=\frac{(3)(0.5)+(2.25)(2.5)+(2.25)(2.5)}{(3)+(2.25)+(2.25)}$$

$$\Rightarrow \bar{y}=1.7$$

Centroid is 1.7in from the left edge of the cross section

Step 2

Find the moment of inertia using the parallel axis theorem for each component and combine

$$I=\frac{1}{12}{bh^3}+Ad^2$$

Moment of inertia for piece 1

$$I=\frac{1}{12}{b_1h_1^3}+A_1d_1^2$$

where:$$ d_1=c_1-\bar{y}=0.5-1.7=-1.2$$

$$I=\frac{1}{12}{(3)(1)^3}+(3)(-1.2)^2$$

$$\Rightarrow I_1=4.57in^4$$

Moment of inertia for piece 2

$$I=\frac{1}{12}{b_2h_2^3}+A_2d_2^2$$

where:$$d_2=c_2-\bar{y}=2.5-1.7=0.8$$

$$I=\frac{1}{12}{(0.75)(3)^3}+(2.25)(0.8)^2$$

$$\Rightarrow I_2=3.1275in^4$$

Moment of inertia for piece 3 is the same as piece 1

$$I_3=I_2=3.1275in^4$$

Total moment of inertia

$$I=I_1+I_2+I_3$$

$$\Rightarrow I=10.825in^4$$

Step 3

The total normal stress at point A is found with:

$$\sigma_x=\frac{P}{A}-\frac{M\bar{y}}{I}$$

where $$M=Pc$$ and c is the distance from point a to the centroid $$\Rightarrow c=3.2in$$

First I'll find the maximum positive tension in the iron hanger which is at the left most point of the iron hanger

$$\sigma_{x_+}=\frac{P_{max}}{A}-\frac{P_{max}c\bar{y}}{I}$$

Solve for $$P_{max}$$ where $$\sigma_{x_+}=+5ksi$$

In this case $$\bar{y_+}=-1.7in$$

Solve total normal stress equation for $$P_{max}$$

$$P_{max}=\frac{\sigma_{x_+}}{\frac{1}{A_{total}}+\frac{c\bar{y_+}}{I}}$$

Plug in known variables

$$P_{max}=\frac{(5)}{\frac{1}{(7.5)}+\frac{(3.2)(1.7)}{(10.825)}}$$

$$\Rightarrow P_{max}=7.86kips$$

In order to find $$P_{max}$$ at the point of maximum compression, the rightmost point is analyzed

So $$\bar{y_-}=4in-1.7in=2.3in$$

Solve for $$P_{max}$$ where $$\sigma_{x_-}=-12ksi$$

$$P_{max}=\frac{\sigma_{x_-}}{\frac{1}{A_{total}}-\frac{c\bar{y_-}}{I}}$$

$$P_{max}=\frac{(-12)}{\frac{1}{(7.5)}-\frac{(3.2)(2.3)}{(10.825)}}$$

$$\Rightarrow P_{max}=-21.95kips$$

The smaller of the two values of $$P_{max}$$ will be the limiting downward force

The largest downward force that can be applied is $$7.86kips$$

When applying an upward force at point A, the iron hanger experiences a tension on the right most side and compression on the left most side. The equations for $$P_{max}$$ remain unchanged except $$\sigma_{x_+}$$ $$\sigma_{x_-}$$ switch places

$$P_{max}=\frac{-\sigma_{x_-}}{\frac{1}{A_{total}}+\frac{c\bar{y_+}}{I}}$$

$$P_{max}=\frac{-(-12)}{\frac{1}{(7.5)}+\frac{(3.2)(1.7)}{(10.825)}}$$

$$\Rightarrow P_{max}= 18.87kips$$

$$P_{max}=\frac{-\sigma_{x_+}}{\frac{1}{A_{total}}-\frac{c\bar{y_-}}{I}}$$

$$P_{max}=\frac{-(5)}{\frac{1}{(7.5)}-\frac{(3.2)(2.3)}{(10.825)}}$$

$$\Rightarrow P_{max}= 9.15 kips$$

The largest upward force that can be applied is 9.15kips

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 6.4
P4.115, Beer 2012

Problem Statement
Solve Prob. 4.114, assuming that the vertical rod is attached at point B instead of point A.

Step 1
First we need to find the centroid and the mass moment of inertia:

$$ y_{bar} = \frac{\sum y A}{\sum A} $$

looking at the cross section as a large rectangle with a smaller rectangle cut from it we find the values:

large rectangle: $$ A = 12 in^{2},  y_{bar} = 2 in $$ smaller rectangle: $$ A = 4.5 in^{2},  y_{bar} = 1.5 in $$

*note* all distances are measured relative to the origin being on the right side of the cross-section $$ y_{bar} = \frac{\sum y A}{\sum A} = \frac{(12in^{2}*2in) - (4.5in^{2}*1.5in)}{7.5 in} = 2.3 in$$

$$ I_{x,rectangle} = \frac{1}{12}b*h^{3} $$

$$ I_{x,rectangle}' = \frac{1}{12}b*h^{3} +Ad^{2}$$

$$ I_{x}' = (\frac{1}{12}3*4^{3}+12*3^2) - (\frac{1}{12}1.5*3^{3}+4.5*.8^2)= 10.825 in^4 $$

Step 2
We find the distance from the centroid to the applied force to be:

d = 2.3 in + 1.5 in = 3.8 in

M = Pd

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

$$ \sigma _{bend}= \frac{Mc}{I} = \frac{-Pdc}{I}$$

$$ \sigma _{c} = \sigma _{o} + \sigma _{bend}= \frac{P}{A} + \frac{Pdc}{I} $$

$$ 5ksi = \frac{P}{7.5} + \frac{P*3.8*2.3}{10.825} $$

Solving for P,gives us $$ P_{max} = 5.31 kips $$

$$ \sigma _{c} = \sigma _{o} - \sigma _{bend} = \frac{P}{A} - \frac{Pdc}{I}$$

$$ -12ksi = \frac{P}{7.5} - \frac{P*3.8*2.3}{10.825} $$

Solving for P,gives us $$ P_{max} = -8.3376 kips $$

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