Beam Loading Example
•
To determine the reaction forces acting at
each support, we must determine the
moments (represented as torques).
•
Static Equilibrium: We can use this fact to
find the conditions for "static equilibrium":
the condition an object is in when there are
forces acting on it, but it is not moving.
•
The conditions for static equilibrium are easy
to state: the sum of the (vector) forces must
equal zero, and the sum of the torques must
equal zero:
–
Σ F = 0 and Σ τ = 0.
•
In this example, b
ecause the beam is in static
equilibrium, the sum of the sum of the forces
in the y

direction is zero and the sum of the
moment torques about one end must also
equal zero.
Forces Acting Upon the Beam
•
The first step is to identify
the forces acting upon the
beam
•
In this example, you have
the downward force of the
block (C) the downward
force of the beam (D), and
the two reaction forces of
the supports (A and B)
•
In this example, the beam
weighs 500 pounds and the
block weighs 750 pounds.
Forces Acting Upon the Beam
•
Because the system is in static
equilibrium, the upward and
downward forces must sum to
be 0
•
Therefore,

A +

B + C + D = 0
•
We’ll use indicate
downward
forces as “+” and upward as “

”
•
Therefore,

A +

B + 500 lbs +
750 lbs = 0
•
We need to determine
the
reaction forces at A and B, but
we have two unknowns and
the block is not at the center
of the beam
Using Moments (Torques)
•
Because the system is in static
equilibrium, the sum of the
moment torques must also
sum to be 0
•
Sum
of the clockwise torques
+ sum of counterclockwise
torques = 0
•
Working from the A support,
we have the following torques:
–
A clockwise
torque at C
–
A clockwise
torque at D
–
A counterclockwise
torque at B
Using Moments (Torques)
•
To calculate the moments, we
need to know the distances
–
(A moment is force X distance)
•
Working from the A support, we
have the following moments:
•
S
M
A
= 0
•
0 = M
C
+ M
D
+ M
B
–
A , there is no moment at A since it
is our analysis
point
–
C = (2 ft X 750 lbs) = +1,500 ft

lbs
–
D = (5 ft X 500 lbs) = +2,500 ft

lbs
–
B =

(10 ft X ?? Lbs)
Using Moments (Torques)
•
Replacing the moments in the equation:
•
S
M
A
= 0
•
0 = M
C
+ M
D
+ M
B
0 = (1,500 ft

lbs)+(2,500 ft

lbs)

(10 ft)(??lbs)
–
Remember, the moment at B is negative because it is counterclockwise.
–
Using algebra, we get:
(
1,500 ft

lbs
)+(2,500
ft

lbs
)=(10 ft)(??
lbs
)
–
We can solve for the force at B
( 4,000 ft

lbs / 10 ft) = 400 lbs
–
So, the reaction force at B is 400 lbs
Beam Forces
•
Going
back to the static equilibrium
formula for forces, we can now
take care of one of the two
unknowns we had in the formula:

A +

B + 500 lbs + 750 lbs = 0

A +

400 lbs + 750 lbs + 500 lbs = 0
Using algebra, we can determine the
reaction force at A

A = 400 lbs +

750 lbs +

500 lbs
A = 850 lbs
Final Diagram
•
If the structural supports at A
and B had a breaking strength
of 900 pounds, what
conclusions would you reach?
•
If the design criteria called for
a safety factor of 4 for the
supports, what conclusions
would you reach?
•
Assuming the forces generated
at A are worst case possible, to
meet the safety factor of 4,
what would the minimum
breaking strength have to be
for the supports?
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