MECHANICS  CASE STUDY SOLUTION

Beam Loading
and Cross Section


Before the stress state at point A can be analyzed for maximum shear and normal
stresses, the maximum bending moment and shear load in the beam needs to be determined.
From the maximum bending moment, the maximum bending stress can be found. Also,
from the maximum shear load, the maximum shear stress can be determined. Both
of these stresses are in the beam oriented coordinate system.
After bending and shear stresses are found, the stress element can be
rotated to given the absolute maximum shear stress. This stress state will most
likely be something other than the natural beam coordinate system.
There are a number of preliminary steps required to find both the beam shear
stress and bending stress. These include determining the
momentshear diagram,
the neutral axis,
moment of inertia, I, and the
first moment of the area, Q.






Moment Shear Diagrams

ShearMoment Diagrams 

To find the maximum bending moment and shear load, the moment and shear equations
can be used. Cutting the structure and summing the forces and
moments, gives,
ΣF_{y} = 0 =
V  50(0.2  x)
V = (10  50x) kN
and
ΣM = 0 =
M  50 (0.2  x) (0.2  x)/2
M = (1  10 x + 25 x^{2})
kNm
The maximum shear and moment will be at the wall, V_{max} = 10 kN and
M_{max} =
1 kNm






Neutral Axis and Moment of Inertia

Neutral Axis Location 

The neutral axis is critical in finding the cross section moment of inertia
and the first moment of the area.
The moment of inertia for full cross section is,
I = I_{1} + y_{1}^{2} A_{1} +
I_{2} + y_{2}^{2} A_{2}
= 40(10)^{3}/12 + (50  38.33 + 10/2)^{2}(40)(10)
+ 10(50)^{3}/12
+ (38.33  50/2)^{2}(50)(10)
= 307,500 mm^{4} = 3.075 × 10^{7} m^{4}






First Moment of the Area, Q

First Moment of Area 1 

The first moment of the area, Q, is needed to determine the shear stress at
point A. The area below or above point A can be used to calculate Q. If the area
above is used, then Q is
Q = y_{1}A_{1} = (50  38.33 + 10/2)(40)(10)
= 16.67 (400)
= 6,668 mm^{3} =
6.668 × 10^{6} m^{3}






Normal and Shear Stress at the Wall
(at Point A)

Distance from NA to Point A 

The maximum moment and shear occur at the wall and thus the maximum stresses
will also occur at the wall. The normal stress at point A is
σ_{b} = My/I
= (1 kNm)(0.01167
m)/(3.075 × 10^{7} m^{4})
= 37.95 MPa
The shear stress at A, is
= 21.68 MPa






Maximum Shear Stress

Stress State at Point A Near Wall
(all arrows pointing in positive directions)
Rotated Stress State at Point A
for Maximum Shear Stress


The bending stress and shear stress at point A is shown on a stress element at the
left. The bending stress is considered to be acting in the x direction. There is
no normal vertical stress, so σ_{y} is zero. The shear stress is acting down on the
right edge of the stress element. Thus, the stress is negative and the shear stress
on the right edge is drawn in the up direction.
The maximum shear stress is
=
± 28.81 MPa
This occurs at an angle of
θ_{τmax} = 20.60^{o}
The rotated normal stresses are equal when the shear stress is a maximum, giving
σ_{x′} = σ_{y′} = (σ_{x} + σ_{y})/2 = 37.95/2 = 18.98 MPa
All rotated stresses are labeled on the stress element at the left. Notice,
the shear stress is actually negative when the shear stress rotation equation
is used. On
the other hand, the maximum shear stress equation above can be either positive
or negative due to
the square root.






Maximum Normal Stresses

Rotated Stress State at Point A
for Maximum Normal Stress
Rotated Stresses as a Function of Angle


The maximum normal stress, or principal stresses σ_{1} and σ_{2},
are
= 18.98 ± 28.81 MPa = 47.79, 9.83 MPa
This occurs at an angle of
θ_{p} = 24.41^{o}
It is interesting to visualize the stresses for any angle by plotting the stresses
as a function of θ. This is shown in the diagram at
the left. Note, the stresses have a period of π, or 180^{o}.
This is due to the symmetric nature of the stresses.



