We will now turn our attention to the forces and torque which develop in a loaded beam. Up to this point we have generally looked at only axial members - members in simple tension or compression; and have considered the forces, stresses, and deformations which occur in such members. We will now look at a particular type of non-axial member - loaded horizontal beams, and will begin the process of determining the forces, toque, stresses, and deformations which occur in these beams. And as we proceed on we will also consider the problem of beam design and/or beam selection. In this first topic, we will focus only on the SHEAR FORCES and BENDING MOMENTS (internal torque) which occur in loaded beams. These quantities are very important, as we shall see, since the axial and shear stresses which will develop in the beam depend on the values of the shear forces and bending moments in the beam.
To understand the shear forces and bending moments in a beam, we will look at a simple example. In Diagram 1, we have shown a simply supported 20 ft. beam with a load of 10,000 lb. acting downward right at the center of the beam. Due to symmetry the two support forces will be equal, with a value of 5000 lb. each. This is the static equilibrium condition for the whole beam.
Next let's examine a section of the beam. We will cut the beam a arbitrary distance (x) between 0 and 10 feet, and apply static equilibrium conditions to the left end section as shown in Diagram 2.. We can do this since as the entire beam is in static equilibrium, then a section of the beam must also be in equilibrium.
In Diagram 2a, we have shown left section of the beam, x feet, long - where x is an
arbitrary distance greater than 0 ft. and less than 10 ft. Notice if we just include the
5000 lb. external support force, the section of the beam is clearly not in equilibrium.
Neither the sum of forces (translational equilibrium), nor the sum of torque (rotational
equilibrium) will sum to zero - as required for equilibrium. Therefore, since we know the beam
section is in equilibrium, there must be some forces and/or torque not accounted for.
In diagram 2b, we have shown the missing force and torque. The 10,000 lb. load which we
originally applied to the beam, and the support force cause internal "shearing
forces" and internal torque called "bending moments" to develop. (We have
symbolically shown these in Diagram 2c.) When we cut the beam, the internal shear force
and bending moment at that point then become an external force and moment (torque) acting
on the section. We have shown these in Diagram 2b, and labeled them V
(shear force) and M (bending moment).
Please note that M is a moment or torque - not a force. It does not appear in the sum of
forces equation when we apply static equilibrium to the section - which will be our next
step.
Equilibrium Conditions:
Sum of Forces in y-direction: + 5000 lb. - V = 0 , solving V = 5000 lb.
Sum of Toque about left end: -V * x + M = 0 , we next substitute
the value of V from the force equation into the torque equation: - 5000 lb. * x +
M = 0 , then solving for M = 5000x (ft-lb.)
These are the equations for the shear force and bending moments for the section of the
beam from 0 to 10 feet. Notice that the internal shear force is a constant value of 5000
lb. for the section, but that the value of the internal torque (bending moment) varies
from 0 ft-lb. at x = 0, to a value of 50,000 ft-lb. at x = 10 ft.
First, however we will finish analyzing our simple beam. So far we have found expressions for the shear force and bending moments (V1 = 5000 lb, M1 = 5000x ft-lb) for section 1 of the beam, between 0 and 10 ft. Now we will look at the next section of the beam. We cut the beam at distance x (ft) from the left end, where x is now greater than 10 ft. and less then 20 ft. and then look at entire section to the left of where we cut the beam (See Diagram 3). Where the beam was cut, we have an internal shear force and bending moment - which now become external. These are shown in Diagram 3 as V2 and M2. (We add the '2', to indicate we are looking at section two of the beam.)
We next apply static equilibrium conditions to the beam section, and obtain:
Equilibrium Conditions:
Sum of Forces in y-direction: + 5000 lb. -10,000 lb. - V = 0 ,
solving V2 = -5000 lb.
Sum of Toque about left end: -10,000 lb * 10 (ft) -V * x (ft) + M = 0
, we next substitute the value of V from the force equation into the torque equation :
-10,000 lb * 10 ft. - (-5000 lb) * x (ft) + M = 0 , then solving for M2 =
-[5000x (ft-lb.) - 100,000] ft-lb.
The two expressions above give the value of the internal shear force and bending moment in
the beam, between the distances of the 10 ft. and 20 ft. A useful way to visualize this
information is to make Shear Force and Bending Moment Diagrams - which are really
the graphs of the shear force and bending moment expressions over the length of the beam.
(See Diagram 4.)
These are a quite useful way of visualizing how the shear force and bending moments
vary through out the beam. We have completed our first Shear Force/Bending Moment Problem.
We have determined the expressions for the shear forces and bending moments in the beam,
and have made accompanying shear force and bending moment diagrams.
Now that we have the general concepts concerning shear forces and bending moments, we want
to step back for a moment and become a little more specific concerning some details, such
as choosing the direction of the shear forces and bending moments.
Continue to:
Topic 4.2: Beams - Shear Force
and Bending Moments II
or Select:
Topic 4: Beams - Table of Contents
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