University of Florida/Egm3520/s13.team5.r5
Problem 5.1 (Problem in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
Two W4 x 13 rolled sections are welded together. Determine that largest couple that can be applied about the zaxis when the assembly is bent.
The steel alloy used has properties and . Use a safety factor of 3.0.
Obtained from Appendix B in Beer, 2012 are the properties of W4x13 Rolledsteel:
Area, = 3.83 in^{2}
Width, = 4.16 in
Moment of Inertia (with respect of yaxis), = 11.3 in^{4}
Solution[edit  edit source]
Step One: Divide assembly in two[edit  edit source]
The first step towards an easy and straightforward solution is the division of the given assembly at the midplane of the yaxis, parallel to the zaxis. 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 AA axis.
Step Two: Solve for moment of inertia about AA[edit  edit source]
The moment of inertia about AA is given by Equation 5.11, derived from the parallelaxis theorem.
(5.11)
where, I_{x} is the moment of inertia through the centroid with respect to the xaxis and d^{2} is the square of the distance from the centroid of the shape to the AA axis. For this particular shape, we can calculate the coordinates of its centroid from symmetry.
Substituting in the values for I_{x}, A, and d^{2} into Equation(5.21), we obtain
Step Three: Extrapolate simplified solution to the entire assembly[edit  edit source]
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
because both sections are identical and symmetrical.
The bending moment, or couple, is given by
(5.12)
for which is the maximum allowable stress which is calculated from and the factor of safety using the following equation
(5.13)
Returning to Eq. 5.12 and substituting in the values for and , which is the width, yields


Problem 5.2 (Problem 4.8 in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
Two W4 x 13 rolled sections are welded together. Determine that largest couple that can be applied about the zaxis when the assembly is bent.
The steel alloy used has properties and . Use a safety factor of 3.0.
Obtained from Appendix B in Beer, 2012 are the properties of W4x13 Rolledsteel:
Area, = 3.83 in^{2}
Width, = 4.060 in
Moment of Inertia (with respect of yaxis), = 3.86 in^{4}
Solution[edit  edit source]
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 zaxis.
The axis AA across the bottom of the top half now corresponds to this slice along the zaxis.
The next step is to calculate the moment of inertia about this new axis, AA. This is done using the parallelaxis theorem, given by:
(5.21)
where, I_{y} is the moment of inertia through the centroid of the shape with respect to the yaxis and d^{2} 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 I_{y}, A, and d^{2} into Equation(5.21), we obtain
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
because both sections are identical and symmetrical.
The bending moment, or couple, is given by
(5.22)
for which is the maximum allowable stress which is calculated from and the factor of safety using the following equation
(5.23)
Returning to Eq. 5.22 and substituting in the values for and , which is the width, yields


Problem 5.3 (Problem 4.13 in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
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 Tbeam's web is to be determined.
Given[edit  edit source]
Area of Tbeam's flange (Area 1):
Area of Tbeam's web (Area 2):
Bending moment about horizontal axis:
Solution[edit  edit source]
Derivation of Force on Shaded Area of Tbeam's Web[edit  edit source]
In order to find the force on the shaded area of the Tbeam's web due to the bending moment, the stress distribution must be used, which is given by the bending stress formula:
(5.31)
By using a small area element , the force on the shaded region can be calculated by:
(5.32)
Which, when integrated, solves for the force we're looking for:
(5.33)
where and 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[edit  edit source]
The Tbeam 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:
Calculation of 2nd Moment of Inertia for Beam Cross Section[edit  edit source]
The 2nd moment of inertia for the entire beam cross section can be calculated using the parallel axis theorem,
(5.34)
where for a rectangle is:
(5.35)
Using (5.34) and (5.35), the 2nd moment of inertia for the Tbeam's flange (top rectangle) can be written as:
(5.36)
which gives:
And the 2nd moment of inertia for the Tbeam's web (bottom rectangle) can be written as:
(5.37)
which gives:
Therefore the total 2nd moment of inertia is the combination of both the flange and web moments:
(5.38)
which gives:
Location of Centroid and Area of the Web's Shaded Region[edit  edit source]
The shaded region of the Tbeam's web has a width of:
and a height of:
Which means the shaded region has an area of:
The distance of the centroid from the neutral axis is:
Therefore, the product of the shaded region's area and centroid distance from the neutral axis is:
Calculation of Force on Shaded Area of Tbeam's Web[edit  edit source]
The force on the shaded region of the Tbeam's web can now be calculated using equation (5.33) by substituting all relevant values:
(5.33)
which gives:
Yielding our final answer,


Problem 5.4 (Problem 4.16 in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
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[edit  edit source]
Maximum stress in compression:
Maximum stress in tension:
Solution[edit  edit source]
Analysis[edit  edit source]
Free body diagram of the beam:
Centroid[edit  edit source]
Splitting the bar's cross section into two rectangles, 1 and 2, with 1 being the upper section and 2 being the lower section
We can use these to find the neutral axis (centroid) for the entire cross section with this relationship
(5.41)
Yielding an answer of
(5.42)
Centroidal Moment of Inertia[edit  edit source]
Using the parallel axis theorem, the centroidal moment of inertia can be represented as
(5.43)
where for a rectangle is
(5.44)
Now, inserting the values for the entire bar
(5.45)
(5.46)
Stress[edit  edit source]
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.
(5.47)
This gives us
For tension:
(5.48)
Which gives us
The smaller of these two values is our answer.

(5.49) 
Problem 5.5 (Problem 4.20 in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
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[edit  edit source]
Stress in tension = 120 MPa
Stress in compression = 150 MPa
Solution[edit  edit source]
For a trapezium Centroidal distance from AD is given by
(5.51)
(5.52)
Area of Trapezium ADEF
(5.53)
(5.54)
Moment of Inertia of rectangle BCEF about AD
(5.55)
(5.56)
Moment of Inertia of triangle ABF about AB (AB=CD)
(5.57)
(5.58)
Base of each Triangle
(5.59)
(5.59)
Moment of Inertia of triangle CDE about CD
(5.510)
(5.511)
Moment of Inertia of trapezium ADEF about AD
(5.512)
(5.513)
(5.514)
Now applying parallel axes theorem of moment of Inertia
(5.515)
Now rearrange the equation to give
(5.516)
(5.517)
(5.518)
Now,
(5.519)
(5.520)
(5.521)
(5.522)
Bottom surface is in compression, for allowable compression stress of 150 MPa
Here
(5.523)
(5.524)
Couple M will be
(5.525)
(5.526)
(5.527)
(5.528)
Both the values given in equations (5.522) and (5.528) Therefore

(5.529) 
Problem 5.6 (Problem 3.53 in Beer, 2012)[edit  edit source]
On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.
Problem Statement[edit  edit source]
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[edit  edit source]
Modulus for aluminum is 3.7*10^6 psi
Modulus for brass is 5.6*10^6 psi
Solution[edit  edit source]
Equilibrium Solution
(3.531)
The compatibility condition
(5.62)
(5.63)
The polar moment of each section
(5.64)
(5.65)
The internal torques
(5.66)
(5.67)
Substituting into the compatibility condition
(5.68)
(5.69)
(5.610)
(5.611)
(5.612)
(5.613)
(5.614)
The max sheer stress in cylinder AB
(5.615)
The max sheer stress in cylinder BC
(5.616)

(5.617) 

(5.618) 
Contributors[edit  edit source]
Team Designee: Daniel Siefman
Table of Assignments
 
Problem Number  Solved by 
Reviewed by 
5.1 
Daniel Siefman & Tim Skankwitz  All 
5.2 
Maria Carasquilla & Gregory Grannell  All 
5.3 
Michael Lindsay & Josh Herrera  All 
5.4 
Tim Shankwitz & Gregory Grannell  All 
5.5 
Andrew Moffatt, Joshua Herrera  All 
5.6 
Phil D Mauro & Andrew Moffat  All 
References[edit  edit source]
Beer, F. P., Johnston, E. R., Jr., DeWolf, J. T., & Mazurek, D. F. (2012). Mechanics of materials (6th ed.). New York, NY: McGraw Hill.