MECHANICS  EXAMPLE



Example 1

Example Graphic 

A simple supported beam needs to support two loads, a point force of 500 lb and a distributed load of 50 lb/ft as shown. Plot the shear and moment over the length of the beam. What is the maximum shear and moment?
In this example, there is a point load and a distributed load. This will require the beam to be sectioned into three segments. Each segment will have a separate moment and shear equation. 





Solution 1

FreeBody Diagram 

The first step in analyzing any beam is to determine the reaction forces at
the supports. Using the free body diagram at the left, the moments can be summed
at point A to give,
ΣM_{A} = 0
12 R_{D}  9 (300)  3 (500) =
0
R_{D} = 350 lb
The reaction at A can be found by summing forces,
ΣF_{y} = 0
R_{A}  500  300 + 350 = 0
R_{A} = 450 lb 



Beam Sections
Section 1 with Sign Convention 

The beam needs to be split into three sections. Each section starts
and ends at a load or support. After the beam is organized into sections, then
each section can be cut and the internal shear and moment determined.
Section 1
The first cut is between points A and B. The location of the cut is not set, and
thus labeled as x. For any location x between x = 0 and x = 3 ft, the shear
and moment are given by
V = 450 0 ≤ x ≤ 3
ft
M = 450x 0 ≤ x ≤ 3
ft 



Section 2 

Section 2
The second cut is done between points B and C. This beam section includes the 500 lb
point force. The section is kept in equilibrium by internal loads, V_{2} and
M_{2}. These two loads can be determined from the equilibrium equations,
giving
ΣF_{y} = 0
450  500  V_{2} = 0
V_{2} = 50 lb 3 ≤ x ≤ 6
ft
ΣM_{cut} =
0
M_{2} + 500 (x  3)  450 x = 0
M_{2} = 1,500  50x 3 ≤ x ≤ 6
ft 



Section 3 

Section 3
The third cut is done between points C and D (left edge). This beam section is the most complex due to the distributed that is cut. Only that part of the distributed load that is on the section is used. The width of the distributed load on the remaining beam section is (x  6) ft.
Just like the other two sections, the unknown internal loads, V_{3} and M_{3}, can be determined from the equilibrium equations, giving
ΣF_{y} = 0
450  500  (x  6) 50  V_{3} = 0
V_{3} = 250  50x lb 6 ≤ x ≤ 9
ft
ΣM_{cut} = 0
M_{3} + 500 (x  3) + [(x6)50] (x6)/2  450 x
= 0
M_{3} = 600 + 250 x  25 x^{2} 6 ≤ x ≤ 9
ft 
Shear and Moment Diagrams 



Plot Shear and Moment Equations
Each of the three segments have different functions for the shear and moment. These can be plotted over each section to give a complete shear and moment diagram. These are shown at the left.
Notice, the point force produces a constant shear in the first two segments. The distributed load cause the shear to vary linearly in the third section.
The moment changes linearly in the first two segments due to the point load.
However, the distributed load produces a quadratic polynomial equation for the
three section. 






Example 2

Skate Board Loading 

A student has decided to design a new type of skateboard made from metal for his school project. The skateboard has two extra metal plates from A to B and C to D. These plates exert distributed loads as shown. His own weight of 140 lb is equally divided between the two leg positions A and C.
What is the maximum moment on the frame and where it is located? (Consider AD as weightless). 





Solution 2

FreeBody Diagram


There will be a reaction at B and D. These can be found by using the static equilibrium equations,
ΣF_{y} = 0
R_{B} + R_{D} = 70 + (40)(0.5) + 70 + (20)(1) = 180 lb
ΣM_{D} = 0
(20)(0.5) + (70)(1) + (20)(2.25) + (70)(2.5) = 2R_{B}
R_{B} = 150 lb
R_{D} = 180  150 = 30 lb 



Internal Loads in Span 1 

Now that the reaction forces are known, the beam can be cut into sections. Notice that there are three spans that need to be checked, A to B, B to C, and C to D.
Section 1
The first section is cut and the left part of the cut beam is used as shown in the diagram. The unknown internal shear V_{1} and moment M_{1} are applied at the cut edge to keep the beam section in equilibrium. These two loads can be determined from the equilibrium equations, giving
ΣF_{y} = 0
70  40x  V_{1} =0
V_{1} = 70  40x lb
ΣM_{cut} = 0
M_{1} + 40x(x/2) + 70x = 0
M_{1} = 20x^{2}  70x ftlb 



Internal Loads in Span 2 

Section 2
Again, cut the beam between B and C, and analyze the left part.
ΣF_{y} = 0
70  (40)(0.5) + 150  V_{2} = 0
V_{2} = 60 lb
ΣM_{cut} = 0
M_{2} + 70x + 20 (x  0.5/2)  150 (x  0.5) = 0
M_{2} = 60x  70 ftlb 



Internal Loads in Span 3 

Section 3
For this section it is convenient to take the right side of the cut. But remember to use the correct positive shear and moment direction.
ΣF_{y} = 0
V_{3}  20 (2.5  x) + 30 = 0
V_{3} = 20  20x lb
ΣM_{cut} = 0
M_{3}  30 (2.5  x) + 20 (2.5  x) (2.5  x)/2 = 0
M_{3} = 10x^{2} + 20x + 12.5 ftlb 



Shear Diagram 

Recall, the shear is the derivative of the moment, dM/dx = V, and thus the moment will be a maximum (or minimum) when the shear is 0. From the shear diagram it is noticed that the shear is zero at x = 0.5 ft and x = 1.5 ft. At those points, the moment is 40 ftlb (x = 0.5) and 20 ftlb (x = 1.5). Hence, maximum moment magnitude is 40 ftlb and it is located at point B.
It should be noted that, for this example the moment diagram was not needed to determine the maximum moment and its location. 






Example 3

Beam with Linear Distributed Load


Find the maximum moment for a simply supported beam with a linear varying distributed load as shown. 



Solution 3

FreeBody Diagram


The first step in analyzing any beam is to determine the reaction forces at the supports. Using the free body diagram at the left, the moments can be summed at point B giving
ΣM_{B } = 0
R_{1}L  0.5 W_{o}L (L/3) = 0
R_{1} = W_{o}L/6
ΣF_{y} = 0
R_{1} + R_{2}  W_{o}L/2 = 0
W_{o}L/6 + R_{2}  W_{o}L/2 = 0
R_{2} = W_{o}L/3




 
To develop the shear force and bending moment equation the beam has to be split at a distance of x from point A. The shear force and bending moment equations are given by
ΣF_{y} = 0
V + W_{o}L/6  0.5 (W_{o}/L) x^{2 }= 0
V = W_{o}L/6  0.5 W_{o}x^{2}/L
ΣM_{cut} = 0
M  W_{o}L/6 (x) + 0.5 (W_{o}x^{2}/2)(2x/3) = 0
M = W_{o}Lx/6  W_{o}x^{3}/(6L)
Recall, the shear, V is equal to the derivative of the moment, dM/dx. So the maximum (or minimum) will occur when the shear is zero.
dM/dx = V = 0
W_{o}L/6  W_{o}x^{2}/(2L) = 0
Substituting the value of x in moment equation maximum moment can be found.



