Van Warren
Warren Design Vision
(c) 1994 All Rights Reserved
A simply supported beam is subjected to a concentrated load somewhere along its span:
Determine the reactions at each of the supports, the shear and bending moment, the deflection along the span, and the slope at the endpoints.
L = beam length
P = the concentrated force
a = x coordinate of applied force
Ra = the reaction at support A
Rb = the reaction at support A
y(x) = deflection as a function of x
V(x) = shear as a function of x
M(x) = bending moment as a function of x
A subscript of '1', refers to interval 0 <= x <= a
A subscript of '2', refers to interval a <= x <= L
Note that the point x = a is shared by intervals 1 & 2.
First we sum forces and moments to deduce the values of the reactions at supports A and B:
In Mathematica÷& notation:
The following expressions can be output to a spreadsheet or program for repetitive evaluation under varying conditions:
We now have the reactions at each support in terms of the load, placement and beam length. Now we need to compute the shear, bending moment, beam deflection and end slopes. Before we get into that however, we need to establish some sign conventions.
Consider the following beam with imaginary cutting sections labeled A' and B':
Consider two beam segments, the segment A-A' and the segment A'-B'.
The external loads impressed on the structure tend to move segment A-A' up and segment A'-B' down. Therefore we record the sign of the shear as positive in the interval 0 <= x <= a, in accordance with common sign convention. For further discussion of this issue see Beer & Johnston, Vector Mechanics For Engineers 1977, Statics & Dynamics, p 273. As we pass the point at which the load is applied, the shear reverses its sense according to the sign convention.
Mathematically we have:
which graphically looks like:
We now consider the bending moment as a function of position. Returning to the segment A-B', we note that the external forces acting on the structure produce tend to bow the cutting section down. This corresponds to a positive sign for the bending moment.
This continues to be true even after we pass the location at which the load is being applied, so the bending moment is positive along the entire span. Solving for the bending:
Solving for the bending:
We insert the boundary conditions to find the constants:
This gives us a complete result for bending, it looks like
The maximum magnitude of M(x) is obtained by substituting 'a' for x in the moment expressions above. This is left as an exercise for the reader. We now will repeat this argument using the symbolic manipulation facilities of Mathematica÷&:
We now assign y deflection coordinates to two intervals along the beam. The y1 coordinate refers to the interval 0 <= x <= a, and the y2 coordinate refers to the interval a < x <= L. To compute spanwise deflection and slope we refer to linear bending theory which (under the assumption of constant stiffness) states that:
Applying this to the y1 interval (0 <= x <= a) and integrating:
Integrating this result once again we have:
Repeating the procedure for the second interval (a < x <= L) we have:
Integrating once more to find deflection in the second interval yields:
We have four equations, and four unknown constants C1, C2, C1' and C2'. The value of the constants can be determined by applying the four boundary conditions:
Evaluating these manually we obtain:
Evaluating these automatically we have:
We evaluate the constants in terms of the reactions:
This requires little effort compared to checking by hand. We proceed to check our deflection and slope equations in a similar effortless manner:
Our only remaining errand is to evaluate the slope expression at the endpoints and to remember that end slopes come out in radians:
![[Graphics:images/SimpleCFgr55.gif]](images/SimpleCFgr55.gif)