
MECHANICS  CASE STUDY SOLUTION

FreeBodyDiagram (FBD) of Ladder


A four foot folding ladder is being considered for a work platform on a sloping roof. The roof angle is 38 degrees and the ladder needs to be horizontal when placed on the roof as shown in the freebodydiagram (FBD) at the left.
There are two loads on the ladder, the person working and the roofing materials. The person is a concentrated force of 150 lb and the materials is a distributed force ranging from 200 to 100 lb/ft. The location of both are indicated in the diagram.
The total moment, shear force and axial force on the top member of the ladder needs to be determined at a location just to the right of the joint.
Distance 's' can be determined from the Law of the Cosines,
s^{2} = 4^{2} + 4^{2}  2(4)(4) cos 104
s = 6.304 ft






Equivalent Loads

Equivalent Forces
Revised FBD of Ladder


The distributed load can be converted into two concentrated loads which will simplify the solution process. The two equivalent loads are,
F_{1} = 0.5 (2 ft) (100 lb/ft) = 100 lb (triangle
area)
F_{2} = (2 ft) (100 lb/ft) = 200
lb (rectangle area)
The location of each concentrated load is at the centroid of their area. For
the triangle, it is (1/3)2 = 2/3 ft from the left edge.
The revised FBD, reflecting the new equivalent forces, is shown at the left.
Important angles and distances are also shown on the diagram. The vertical
and horizontal distance between points A and B can be determined as
(AB)_{x} = 4 sin14 = 0.9677 ft
(AB)_{y}= 4 cos14 = 3.881 ft




Reactions


Before the internal loads can be determined, the external reactions need to
be found. In this case, there are three unknowns, so the three equilibrium equations
for 2D can be applied to give,
ΣM_{A} = 0
6.304R  (2/3 + 0.9677)(100)  (1 + 0.9677)(200)

150(3 + 0.9677) = 0
R = 182.8 lb
ΣF_{y} = 0
A_{y}  200  150  100 + R cos38 = 0
A_{y} = 450  182.8 cos38
A_{y} = 306.0 lb
ΣF_{x} = 0
A_{x}  R sin38 = 0
A_{x}  182.8 sin38 = 0
A_{x} = 112.5 lb






Section Cut at Specified Location

FBD of Left Section 

The problem asks for the moment, M, shear, V, and axial load, A, just left
of the middle joint, B. This can be done by cutting the ladder at that location
and analyzing either the left or right section. The FBD of the left section is
shown at the left. This section must be in
static equilibrium, just like the full structure. Thus, the three equilibrium
equations can be used to solve for the three unknowns, M, V, and A.
ΣF_{x} = 0
A
+ 112.5 = 0
A = 112.5 lb (compression)
ΣF_{y} =
0
V + 306.0 = 0
V = 306.0 lb
ΣM_{cut} =
0
M + 112.5 (3.881)  306.0 (0.9677) = 0
M = 140.5 lbft






Alternate Solution Method  Check

FBD of Right Section 

As mentioned previously, either side of the cut can be used to determine the
moment, M, shear, V, and axial load, A. As a check, the right section will now
be analyzed and compared to the previous solution. Of course, the two solutions
should match but many times they do not due to calculation errors.
The FBD of the right section is shown at the left. Notice, the loads M, V, and
A are in the opposite direction as the previous analysis since the opposite section
is used. Summing the forces and the moments give,
ΣF_{x} = 0
A  182.8 sin38 = 0
A = 112.5 lb (compression)
ΣF_{y} = 0
V  100  200  150 + 182.8 cos38 = 0
V = 360.0 lb
ΣM_{cut} =
0
M  100(2/3)  200(1)  150(3) + 4(182.8) cos38
M = 140.5 lbft
Notice, all three values match the previous calculations, as they should.




Internal Moment of Ladder as
a Function of Location 

Change in Moment Across Ladder


While not requested in this case, it is interesting to visualize the change
in internal moment if the cut was moved to other locations. The graph at the
left indicates the internal moment magnitude as a function of the cut location,
s. The starting point is at A, then goes to B and then to C.
Notice the maximum is not at joint B, but at the point where the person
is standing. However, joint C does have a large moment and since it is a joint,
it would probably fail before the straight section where the person is standing.




