|
Say I have
the following beam:
I want to
calculate the reactions at the supports. I do so by following
these steps:
Step
1) Draw the Free Body
Diagram (FBD)
A) Decide
upon a sign convention. It does not matter what sign convention
you choose, as long as you are consistent throughout your calculations.
B) Draw beam
with support reactions acting on it. Note that the arrows can
be oriented in any direction, because they are unknown at this
stage. When they are calculated, the correct orientation will
become obvious.
C) Draw the
additional forces acting on the beam.
(a) The
total contribution of the uniform load must be calculated.
| Total
Force |
=
(Uniform Load)(Length of beam it acts upon) |
|
=
(20 lb/ft)(10 ft) |
|
=
200 lb |
(b) The
location where this total load acts on the beam must be determined.
| XTotal
Force |
=
(Starting Point of Uniform Load) +
(Length of beam it acts upon)/2 |
|
=
0 + (10 ft)/2 |
|
=
5 ft |
(c) Now
the FBD can be completed.
Step
2) Apply equilibrium
equations
A) The sum
the forces in the x-direction = 0
| -Fx1
+ (-100 lb) = 0 |
Therefore, Fx1 = -100
lb |
This means
that Fx1 is force of 100 lb acting in the positive
x-direction.
B) The sum
the forces in the y-direction = 0
| Fy1
+ (-200 lb) + (-Fy2) = 0 |
Therefore, Fy1 = Fy2
+ 200 lb |
But,
we still don’t know Fy2.
C)
The sum of the moments about any point = 0. (Remember
moment equals force times distance!)
(a) Choose
a point O.
(b) Sum
moments
-Fx1(0
ft)+ Fy1(0 ft) + (-200 lb)(5 ft) +
(-Fy2)(10 ft) + (-100 lb)(0 ft) = 0 |
| Therefore,
(10 ft)Fy2 = -1000
ft-lb which gives us Fy2
= -100 lb |
This means
that Fy2 is a force of 100 lb acting in the positive
y-direction. We can now plug this into the previous equation
to find Fy1.
Therefore,
| Fy1
|
=
Fy2 + 200 lb |
|
=
-100 lb + 200 lb |
|
=
100 lb |
This means
that Fy1 is a force of 100 lb acting in the positive
y-direction.
Step
3) Draw final diagram
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