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

Report 5

## Problem 5.1[edit | edit source]

P4.7, Beer 2012

### Problem Statement[edit | edit source]

Two W4x13 rolled sections are welded together as shown. For the steel alloy used: , and a factor of safety of 3.0

### Objective[edit | edit source]

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

### Solution[edit | edit source]

### Step 1[edit | edit source]

Draw dimensions from appendix C.

### Step 2[edit | edit source]

From appendix C for W4x13:

The area is equal to

The moment of inertia about x is equal to

The base is equal to

### Step 3[edit | edit source]

The parallel axis theorem gives us the following

being the moment about the neutral axis

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

Since there are two sections and the moment of inertia of the two sections about the neutral axis is

### Step 4[edit | edit source]

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

### Step 5[edit | edit source]

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

### Honor Pledge[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 5.2[edit | edit source]

P4.8, Beer 2012

### Problem Statement[edit | edit source]

Two W4x13 rolled sections are welded together as shown. For the steel alloy used: , and a factor of safety of 3.0

### Objective[edit | edit source]

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

### Solution[edit | edit source]

### Step 1[edit | edit source]

Draw dimensions from appendix C.

### Step 2[edit | edit source]

From appendix C for W4x13:

The area is equal to

The moment of inertia about y is equal to

The base is equal to

### Step 3[edit | edit source]

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

### Step 4[edit | edit source]

The parallel axis theorem gives us the following

being the moment about the neutral axis

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

Since there are two sections and the moment of inertia of the two sections about the neutral axis is

### Step 5[edit | edit source]

The largest distance from from the centroid to either side is

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

### Honor Pledge[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 5.3[edit | edit source]

P4.13, Beer 2012

### Problem Statement[edit | edit source]

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

### Objective[edit | edit source]

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

### Solution[edit | edit source]

### Step 1[edit | edit source]

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:

### Step 2[edit | edit source]

We have

and

the centroidal Moment of Inertia is

then

### Honor Pledge[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 5.4[edit | edit source]

P4.16, Beer 2012

### Problem Statement[edit | edit source]

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

### Objective[edit | edit source]

Determine the largest couple M that can be applied to the beam for

### Givens[edit | edit source]

b = 40mm

s = 15mm

d = 30mm

h = d-s = 15mm

t = 20mm

=?

=?

### Solution[edit | edit source]

### Step 1[edit | edit source]

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

### Step 2[edit | edit source]

Now We solve for

### Step 3[edit | edit source]

Now we must solve for the Moment of Inertia of the T Shape:

### Step 4[edit | edit source]

We can calculate the maximum tensile strength, given that our maximum compression stress is 30Mpa.

### Step 5[edit | edit source]

Since , the maximum stress is seen through compression. Therefore, we will use that compression stress in the elastic flexural formula:

### Step 6[edit | edit source]

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

### Honor Pledge[edit | edit source]

## Problem 5.5[edit | edit source]

P4.20, Beer 2012

### Problem Statement[edit | edit source]

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

### Objective[edit | edit source]

Determine the largest couple M that can be applied.

### Solution[edit | edit source]

#### Step 1[edit | edit source]

The centroid of a trapezoid is given by

where a = 80 mm, b = 40 mm, and h = 54 mm

so

#### Step 2[edit | edit source]

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[edit | edit source]

The moment of inertia of the triangle is given by:

The Area of the triangle is:

centroid of a triangle is : y =

so dy = 36mm - 30mm = 6mm

#### Step 4[edit | edit source]

The moment of inertia of the rectangle is given by:

The Area of the rectangle is:

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

so dy = 30 mm - 27 mm = 3 mm

#### Step 5[edit | edit source]

#### Step 6[edit | edit source]

Applying the Elastic fexural formula to get:

Looking at the bottom half of the beam gives us c = 30 mm and

Looking at the top half of the beam gives us c = 24 mm and

The larges couple M is felt by the bottom half

### Honor Pledge[edit | edit source]

## Problem 5.6[edit | edit source]

P3.53, Beer 2012

### Problem Statement[edit | edit source]

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 for aluminum and for brass.

### Objective[edit | edit source]

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

### Solution[edit | edit source]

**FBD**

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

**Given**

#### Step 1[edit | edit source]

In order to find the max shearing stress, we need to determine the Torques at point A and C

#### Step 2[edit | edit source]

Find the moment of inertia in each cylinder

#### Step 3[edit | edit source]

Find the max sheer stress in each cylinder

### Honor Pledge[edit | edit source]

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