In this tutorial we will:
In the tutorials for a simply supported beam with point loading and a cantilever beam with uniform distributed load we developed the shearing force and bending moment diagrams by calculating V and M at specific points x along the beam from first principles using free body diagrams.
In this example we express V and M as mathematical functions of x from which we can obtain V and M at any point x along the beam.
Consider this example of a simply supported beam 12 m long with variable distributed load w. The load is a linear distribution from w = 12 kN/m at x = 0 to w = 0 at x = 12.
The first step is to express w as a function of x, in this case a linear function with slope = -1 and intercept on the y axis = 12.
This is equivalent to a linear function w = (12 - x)
Next step is to find the equivalent total load and its point of application. We derive the equivalent total load through the following integral.
This load acts at a point through the centroid of the load triangle which is 1/3 from the base at the LH end (a centroid position can also be calculated for non-linear load functions).
Reactions R1 and R2 are then determined using Σ forces = 0 and Σ moments = 0.
Σ forces = 0 gives R1 + R2 = 72 --------- (i)
Σ moments = 0 at x = 0 gives
(72 x 4) - (R2 x 12 ) = 0 Gives R2 = 24 kN
Substitute R2 in (i) gives R1 = 48 kN
Next we determine the function for the shearing force V(x) from the indefinite integral
To find the constant of integration C consider V(x) at x = 0
Substituting x = 0 in V(x) gives V(0) = C
We know V(0) = R1 = 48 kN
Thus C = 48
Calculated values of V and the corresponding shearing force diagram are
shown below. The diagram plots V at x = 0, x = 2, x = 4, x = 6, x =
8, x = 10 and x = 12 . The true curve is a continuous
quadratic function.
x (m) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
V (kN) | 48 | 36.5 | 26 | 16.5 | 8 | 0.5 | -6 | -11.5 | -16 | -19.5 | -22 | -23.5 | -24 |
for which the solution is x = 5.06
To obtain a function M(x) for the bending moment we use the relationship
To find the constant of integration C consider M(x) at x = 0
Substituting x = 0 in M(x) gives M(0) = C
We know M(0) = 0 Thus C = 0
Calculated values of M (rounded to integer
values) and the
corresponding bending moment diagram are
shown below. The diagram plots V for integer values of x in
the table. The true curve is a third order function of x.
x (m) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
M (kNm) | 0 | 42 | 73 | 95 | 107 | 111 | 108 | 99 | 86 | 68 | 47 | 24 | 0 |
it follows that the maximum value of M(x) occurs where V(x) = 0 (x = 5.06)
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