Beam Loading Example

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15 Νοε 2013 (πριν από 3 χρόνια και 7 μήνες)

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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?