Strength of Materials/Loading of Beams< Strength of Materials
One of the areas where solid mechanics as discussed in this book is most effective is in the case beam loading. The loads on a beam can be point loads, distributed loads, or varying loads. There can also be point moments on the beam. The beam itself is supported at one or more points. The conditions at the support depend on the kind of support used. If the support is a roller, it can only have a reaction perpendicular to the motion of the roller. If the support is a pin, it cannot carry a moment. If the support is fixed, then it can have a reaction in any direction and support a moment as well.
In the above image, a simple beam is loaded at the center by a load P. It has a pinned contact at one end, and a rolling contact at the other end.
The above image shows an ideal moment acting at the center of a beam. An ideal moment is one which is not associated with a force.
The general method is analyzing beam problems is to find the loads, reactions, and moments, and come up with the values for the loads and moments at each section. This, in general will be a piecewise function of the distance along the beam.
For loads, we set up axes and this decides the sign of the force. For moments, we set the convention that the clockwise moment is positive.
1. Consider the case of the beam discussed earlier shown above.
The beam has reactions R1 and R2 acting on each of the supports.
R1 + R2 = P
Applying symmetry, we have,
R1 = R2
R1 = R2 = P/2
The above image shows the shear diagram for this problem. Note that the positive and negative directions are conventions, but it is important to choose one direction for positive shear and stick to it. As can be seen, the shear value changes at the point of application of the load.
The above image shows the moment diagram for the beam. The moment varies linearly from the supports to the middle. (The value at the middle should be PL/4). Note that the moment is maximum at the middle and zero at the ends, so that the effect of the moment is maximum at the center. In later chapters we will see that the stresses and strains due to moments are the most important ones for beams.
1. In the above beam, find the reactions in the supports and the shear force at a position x. Also, find the moment at that point.
2. The above beam shows loading by two separate point loads. Find the shear and moment at location x from one end of the beam.
3. Find the shear and moment at each point along the beam with a point load and moment acting at two different points. How can a point moment occur in practice?
Calculus for Solving Beam ProblemsEdit
Methods of calculus can be used to deal with continuous load functions. However these methods can be extended to point loads and moments by the use of the Dirac delta function.
The above shows a beam with uniform load per unit length w. Such loads are used to model the self weight of the beam where it acts uniformly throughout its length. This can also be used for the loading of, say, a bridge due to all the vehicles on it. The individual point loads acting through the tires can be modeled as a continuous load if the number of vehicles is large.
Consider a beam with load per unit length q(x) for any point x along its length. Consider an elemental length dx of the beam. Applying force equilibrium for this segment, we have,
V + dV = q dx + V
where V is the shear at the point x. Thus, we have the relation
dV/dx = q
Applying moment equilibrium to the same segment, we have,
M + dM − V dx − M − q dx dx/2 = 0
Neglecting the second order terms in dx, we have,
dM/dx = V
d2M/dx2 = q
The above differential equations can be integrated with appropriate boundary conditions to get the shear and moment at each point.
Consider a beam of length L supported at its ends by two pins, with a uniform load per unit length of w.
The total load on the beam is thus wL. Thus, the reaction at each support is wL/2.
We have, the general relation for shear
The shear at origin is just the reaction at that point (=wL/2). If we take the vertical direction as the positive direction, we have, the shear at the origin is wL/2. Also, q = −w. Thus, we have, V = −wx + wL/2 for any point x.
The Moment at the centre should be wL^2/4
Similarly, we have the moment M
It is easy to see that the moment at the origin is zero. Thus, we have M = w x (L − x)/2.
Calculus for Point LoadsEdit
The above works well for well for continuous, smooth load functions. But in real world situations, we have to deal with point loads and moments. Thus we use the Dirac functions for a point load P at location a as
and for a point moment M at location a as
Now the previously stated equations can be used under the rules for the Dirac function. The function itself is defined as
Consider a beam of length L with supports at both ends and a point load P at the center. From the Dirac definition, we have,
The shear at one end is just the reaction P/2. The shear at any point is given by
The moment at one end is zero. The moment at any point is given by
Applying the definition of the Dirac function, we have, the shear
Note that the above is a special case of the general method called Laplace Transforms.
1. Find the shear and moment in the above beam using Dirac methods.
2. The beam shown above has two loads which can be modeled as shown. Find the shear and moment at any point along the beam.
3. Consider the beam shown above with an overhang. Find the shear and moment at points along the axis.
Stresses and Strains in BeamsEdit
Given the loads and moments at each cross section, we can calculate the stress and strain at each location.
Consider the bending of a slender beam (one for which the cross section is much smaller than the length). A moment acting on the beam causes a deformation called flexure. Let the radius of the osculating circle of the beam be ρ. Consider an elemental length ds in the neutral plane (for which the deformation is zero). This element subtends an angle θ at the center of curvature, so that
ds/dθ = ρ
If we move a distance y along the radius, we have the length of the arc subtended would be (ρ − y) dθ. Thus the elemental extension would be y dθ. For a small enough curvature, we have the distance along the neutral plane would be the same as the initial undeformed length, dx. This gives us the axial strain at any point x as
εx = y/ρ
Also, using the Hooke's law, we have,
σx = E εx = Ey/ρ
Now, we know that a pure moment M acts on the beam, so that there is no axial force. Or,
∑ Fx = 0
∫ σx dA = 0
∫ y dA = 0
The above equation gives us the location of the neutral plane.
Further, applying the moment conservation part of the equilibrium relations, we have,
M − ∫ σx dA y = 0
Now, we know,
σx = Ey/ρ
I = ∫ y2 dA
where I is the moment of inertia, so that
M = E/ρ I
E/ρ = M/I
Thus, we have the expression for the stress due to a pure moment as
σx = My/I
Suppose the maximum value of y is c (the distance from the neutral plane). Then, we have,
σmax = Mc/I
The quantity I/c is called the section modulus of a beam and is denoted by S.
σmax = M/S
Beams with Arbitrary MomentsEdit
So far we have considered beams which were acted upon by moments in the Mz. Suppose there is another moment in the My direction. Note that a moment in the Mx direction would simply be a torsional effect as the axis is in the x direction. Now, it is easy to see that the combination of moments is, in fact, equivalent to a moment acting on a beam of arbitrary cross section. It is fairly straightforward to show that, in this case, the stress at any point (y, z) on the face of the section for the beam is a function of Iz, Iy, and Iyz.
Shear in BeamsEdit
Earlier, we saw the methods to calculate the shear forces on a beam. Now we can analyze the stresses due to shear forces, like we did for stresses due to bending moments. We showed that the shear force V is given by V = dM/dx, where M is the moment acting at the point x.