Various Types of Beam Loading and Support Beam

Various Types of Beam Loading and Support • Beam - structural member designed to support loads applied at various points along its length. • Beam can be subjected to concentrated loads or distributed loads or combination of both. • Beam design is two-step process: 1) determine shearing forces and bending moments produced by applied loads 2) select cross-section best suited to resist shearing forces and bending moments 7 - 1

Various Types of Beam Loading and Support • Beams are classified according to way in which they are supported. • Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically indeterminate. 7 - 2

Shear and Bending Moment in a Beam • Wish to determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads. • Determine reactions at supports by treating whole beam as free-body. • Cut beam at C and draw free-body diagrams for AC and CB. By definition, positive sense for internal force-couple systems are as shown. • From equilibrium considerations, determine M and V or M’ and V’. 7 - 3

Shear and Bending Moment Diagrams • Variation of shear and bending moment along beam may be plotted. • Determine reactions at supports. • Cut beam at C and consider member AC, • Cut beam at E and consider member EB, • For a beam subjected to concentrated loads, shear is constant between loading points and moment varies linearly. 7 - 4

Step 1: After calculating the reactions at A and B, start the Shear Force Diagram at the first value of the force acting on the beam. In this case it is a +10 k. N due to the reaction at point A: 7 - 5

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step 2: Keep moving across the beam, stopping at every load that acts on the beam. When you get to a load, add to the Shear Force Diagram by the amount of the force. In this case we have come to a negative 20 k. N force, so we will minus 20 k. N from the existing 10 k. N. i. e. 10 k. N - 20 k. N = -10 k. N. 7 - 7

Step 2 (repeated): Moving across the beam again, we come to another force; a positive 10 k. N reaction at support B. Again, add this +10 k. N to the shear force diagram (which is currently at -10 k. N) which will bring us to a shear force of 0. Since we are at the end of the beam, we will go no further and we have our final Shear Force Diagram (SFD): 7 - 8

Bending Moment Diagram of simply supported beam 7 - 9

From left to right, make "cuts" before and after each reaction/load TO calculate the bending moment of a beam, we must work in the same way we did for the Shear Force Diagram. Starting at x = 0 we will move across the beam and calculate the bending moment at each point. Cut 1 Make a "cut" just after the first reaction of the beam. Cut 1 There is only one force applied to the beam The bending at this point increases as you move across the beam 7 - 10

So, when we cut the beam, we only cosider the forces that are applied to the left of our cut. In this case we have a 10 k. N force in the upward direction. Now as you recall, a bending moment is simply the force x distance. So as we move further from the force, the magnitude of the bending moment will increase. We can see this in our BMD. The equation for this part of our bending moment diagram is: -M(x) = 10(-x) M(x) = 10 x Cut 2 This cut is made just before the second force along the beam. Since there are no other loads applied between the first and second cut, the bending moment equation will remain the same. This means we can calculate the maximum bending moment (in this case at the midpoint, or x = 5) by simply substituting x=5 into the above equation: 7 - 11

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Cut 3 This cut is made just after the second force along the beam. Now we have TWO forces that act to the left of our cut: a 10 k. N support reaction and a 20 k. N downward acting load. So now we must consider both these forces as we progress along our beam. For every metre we move across the beam, there will a +10 k. Nm moment added from the first force and -20 k. Nm from the second. So after the point x=5, our Bending Moment Equation becomes: M(x) = 50 +10(x-5) - 20(x-5) M(x) = 50 -10(x-5) for 5 ≤ x ≤ 10 NOTE: The reason we write (x-5) is because we want to know the distance from the pt x=5 only. Anything before this point uses a previous equation. 7 - 13

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Cut 4 Again, let's move across to the right of our beam and make a cut just before our next force. In this case, our next cut will occur just before the reaction from Right Support. Since there are no other forces between the support and our previous cut, the equation will remain the same: M(x) = 50 -10(x-5) for 5 ≤ x≤ 10 And let's substitute x=10 into this to find the find bending moment at the end of the beam: M(x) = 50 - 10(10 -5) = 0 k. Nm This makes perfect sense. Since our beam is static (and not rotation) it makes sense that our beam should have zero moment at this point when we consider all our forces. It also satisfies one of our initial conditions, that the sum of moments at a support is equal to zero. NOTE: If your calculations lead you to any other number other than 0, you have made a mistake! 7 - 15

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