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: ${\sigma }_y=36ksi$, ${\sigma }_U=58ksi$ and a factor of safety of 3.0

File:P4.7.png

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.
File:P4.7-1.png

Step 2

From appendix C for W4x13:

The area is equal to $A=3.83i{n}^2$

The moment of inertia about x is equal to $I_x=11.3i{n}^4$

The base is equal to $b=4.06in$

Step 3

The parallel axis theorem gives us the following

$I_x=I_{x^{\prime }}-A(x{)}^2$

$I_{x^{\prime }}$ being the moment about the neutral axis

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

$I_{x^{\prime }}=I_x+A(x{)}^2$

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

$I_s=2[I_x+A*(2.08{)}^2]=2[11.3+3.83*(208{)}^2]=55.7i{n}^4$

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

$M_{max}=\frac{{\sigma }_m*I_s}{c}=\frac{19.33*55.7}{4.16}=259kip*in$

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: ${\sigma }_y=36ksi$, ${\sigma }_U=58ksi$ and a factor of safety of 3.0

File:P4.8.png

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.
File:P4.8-1.png

Step 2

From appendix C for W4x13:

The area is equal to $A=3.83i{n}^2$

The moment of inertia about y is equal to $I_Y=3.86i{n}^4$

The base is equal to $b=4.06in$

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

$I_s=2[I_y+A*(2.03{)}^2]=2[3.86+3.83*4.12]=39.29i{n}^4$

Step 5

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

$M_{max}=\frac{{\sigma }_a*I_s}{c}=\frac{19.33*39.29}{4.06}=187.1kip*in$

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.
File:P4.13.png

Objective

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

Solution

File:P4.13-1.png

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:

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

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

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

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

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

Step 2

We have

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

and $\overline{y}=0.045m$

the centroidal Moment of Inertia is

$I=\sum (I_x+A{d}^2)$

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

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

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.

File:P4.16.png

Objective

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

Givens

File:4.16 FBD.jpg

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:
$I=\frac{1}{3}[t*\stackrel{3}{\stackrel{‾}{y_2}}+b*\stackrel{3}{\stackrel{‾}{y_1}}-(b-t)(\overline{y_1}-s{)}^3]$

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

Step 4

We can calculate the maximum tensile strength, given that our maximum compression stress is 30Mpa.
${\sigma }_{Tmax}=\frac{\overline{y_1}}{\overline{y_2}}*30Mpa=\frac{16.59}{23.41}*30Mpa=21.26Mpa$

Step 5

Since ${\sigma }_{Tmax}<24Mpa$, 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:

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

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.
File:P4.20.png

Objective

Determine the largest couple M that can be applied.

Solution

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

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

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:

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

The Area of the triangle is:

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

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:

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

The Area of the rectangle is:

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

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

so dy = 30 mm - 27 mm = 3 mm

Step 5

$I_{xx}=2(I_{triangle}+Ad{y}^2)+(I_{rectangle}+Ad{y}^2)$

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

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 $\sigma =150MPa$

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

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

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

The larges couple M is felt by the bottom half

$M_{max}=3748.1N*m$

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 $3.7*10^{6}psi$ for aluminum and $5.6*10^{6}psi$ for brass.

File:P3.53.png

Objective

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

Solution

FBD

File:5.6 FBD.PNG

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
$\sum M_x=0$
$\Rightarrow T_{AB}+T_{BC}-T=0\Rightarrow T_{AB}+T_{BC}=12.5kips\cdot in$

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

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

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

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

Step 2

Find the moment of inertia in each cylinder

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

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

Step 3

Find the max sheer stress in each cylinder

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

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

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

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.