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© D.J.DUNN 1
MECHANICS OF SOLIDS - BEAMS TUTORIAL 3 THE DEFLECTION OF BEAMS
This is the third tutorial on the bending of beams. You should judge your progress by completing the self assessment exercises. On completion of this tutorial you should be able to solve the slope and deflection of the following types of beams.
A cantilever beam with a point load at the end.
A cantilever beam with a uniformly distributed load.
A simply supported beam with a point load at the middle.
A simply supported beam with a uniformly distributed load.
You will also learn and apply Macaulay’s method to the solution for beams
with a combination of loads. Those who require more advanced studies may also
apply Macaulay’s
method to the solution of ENCASTRÉ.
It is assumed that students doing this tutorial already know how to find the bending moment in various types of beams. This information is contained in tutorial 2.
© D.J.DUNN 2
DEFLECTION OF BEAMS 1. GENERAL THEORY
When a beam bends it takes up various shapes such as that illustrated in figure 1. The shape may be superimposed on an x
–
y graph with the srcin at the left end of the beam (before it is loaded). At any distance x metres from the left end, the beam will have a deflection y and a gradient or slope dy/dx and it is these that we are concerned with in this tutorial. We have already examined the equation relating bending moment and radius of curvature in a beam, namely
R EIM
M is the bending moment. I is the second moment of area about the centroid. E is the modulus of elasticity and R is the radius of curvature. Rearranging we have
EIMR 1
Figure 1 illustrates the radius of curvature which is defined as the radius of a circle that has a tangent the same as the point on the x-y graph. Figure 1 Mathematically it can be shown that any curve plotted on x - y graph has a radius of curvature of defined as
2322
dxdy1dxydR 1
© D.J.DUNN 3
In beams, R is very large and the equation may be simplified without loss of accuracy to
22
dyxdR 1
hence
EIMdyxd
22
or
..(1A)..........dyxdEIM
22
The product EI is called the flexural stiffness of the beam. In order to solve the slope (dy/dx) or the deflection (y) at any point on the beam, an equation for M in terms of position x must be substituted into equation (1A). We will now examine this for the 4 standard cases.
© D.J.DUNN 4
2. CASE 1 - CANTILEVER WITH POINT LOAD AT FREE END.
Figure 2 The bending moment at any position x is simply -Fx. Substituting this into equation 1A we have
......(2B)..........BAx
6Fx-EIy get weandagain Integrate
(2A)..........A.........2FxdxdyEI get weandwrtx Integrate
FxdxydEI
3222
A and B are constants of integration and must be found from the boundary conditions. These are at x = L, y = 0 (no deflection) at x = L, dy/dx = 0 (gradient horizontal) Substitute x = L and dy/dx = 0 in equation 2A. This gives
D)........(2..........3FL-2xFL6FxEIyC)........(2..........2FL2FxdxdyEIareequationscompletetheand2Band2A equationsinto
3FL-Band2FLA substitute3FL-Bhence B
2FL6FLEI(0)getweand2Bequation ntoLx and0y,
2FLA substitute2FLA hence A
2FLEI(0)
3232232333222
The main point of interest is the slope and deflection at the free end where x=0. Substituting x= 0 into (2C) and (2D) gives the standard equations. Slope at free end
E)........(2..........2EIFLdxdy
2
Deflection at free end
F)........(2..........3EIFLy
3

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