MECHANICS  CASE STUDY

Example Graphic
Beam Model with Loading


Introduction


The trailer is 53 ft long and carries 79,500 lb. Since the load is assumed
to be evenly distributed across the trailer bed, the load intensity, w, is
w = (79,500 lb)/(53 ft) = 1,500 lb/ft =
1.5 kip/ft
The trailer is support by two sets of wheels which can be modeled as simple
pin joints as shown at the left. Joint A is at the left edge and joint B is at
the center of the back wheels. Both of the support joints are approximate.





Free Body Diagram with Reactions

Free Body Diagram


The basic trailer can be modeled with a free body diagram to assist in finding
the reaction forces. There will be a reaction at joint A and B. These can be
found by using the static equilibrium equations,
ΣM_{A} = 0
39 R_{B}  26.5 (79,500) =
0
R_{B} = 54,020 lb = 54.02 kip
The reaction at A can be found by summing forces,
ΣF_{y} = 0
R_{A} + 54.02  79.5 = 0
R_{A} = 25.48 kip




Beam Sections 

Sections


Now that the reaction forces are known, the beam can be cut into
sections. This beam has one continuous load and two supports. Since one of the
supports, the right one, is not at the beam edge, this will cause a discontinuity
in the moment and shear diagrams. The beam will need to be analyzed at two places,
once in each of the sections shown in the diagram. 





Section 1)

Section 1 Cut


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
25.48  1.5x  V_{1} = 0
V_{1} = 25.48  1.5x 0 ≤ x ≤ 39
ft
ΣM_{cut} =
0
M_{1} + (x/2) (1.5x)  25.48x = 0
M_{1} = 25.48x  0.75x^{2} 0 ≤ x ≤ 39
ft






Section 2)

Section 2 Cut 

The second cut is done between joint B and the right edge of the beam. This
time, the right beam section will be analyzed. Again, the unknown internal
shear V_{2} and moment M_{2} are applied at the cut edge. Note,
that the direction of the internal loads are opposite from section 1 cut. This
is because the opposite beam section (the right side) is used. This follows
the sign
convention
specified in the previous section.
Applying the equilibrium equations gives,
ΣF_{y} = 0
V_{2}  1.5 (53  x) = 0
V_{2} = 79.5  1.5x 39 ≤ x ≤ 53
ft
ΣM_{cut} =
0
M_{2}  [(53  x)/2] [1.5 (53  x)]
= 0
M_{2} = 2,107 + 79.5x  0.75x^{2} 39 ≤ x ≤ 53
ft






Plot Shear and Moment Equations

Shear and Moment Diagrams 

Each of the two 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.
With many shear and moment diagrams, a maximum is needed. In this problem,
the first beam section increases and then decreases. To find the location of
the maximum, equate the first derivative with respect to the locations, x, to
zero. This gives,
dM_{1}/dx = 0
25.48  1.5x = 0
x = 16.99 ft
The maximum moment at 16.99 ft is 216.4 kipft which is a large moment. The
moment is positive so the beam will be bent downward. This moment does exceed
the maximum allowed of 200 kipft, so failure would be expected.




Truck Failed Near the Location of
the Maximum Moment


Discussion


It is interesting to note that the maximum moment from the analysis
agrees well with the location where the trailer failed.
The failure can be due to other issues such as road quality and exact load
distribution (it was assumed constant). Also the boundary conditions were
assumed pinned which is a simplification to allow analysis.



