For the loaded beam shown in the figure, the correct shear force diagram is
The shear force in the span BC will be zero. The shear force in the span AB and CD will be of opposite sign.
For the simply supported beam, shown in the figure below at what distance from the support A is the shear force zero?
Let the point of zero shear force occur at section X-Xfrom left support A
∴ Shear force at X-X
∴ For zero shear force we have,
For the beam shown ih the given figure, the maximum positive bending moment is equal to the maximum negative bending moment. The value of L1 is
The BM can be found from the area of SFD.
The shear force diagram is
Maximum negative, bending moment,
Maximum positive, bending moment,
For M1 = M2
The maximum bending moment due to a moving load on a fixed ended beam occurs .
If a beam is subjected to a constant bending moment along its length then the shear force will
If M is constant then,
S.F = 0, as
S.F = dM/dx
If the shear force acting at every section of a beam is of the same magnitude and of the same direction, then it represents a
The shear force diagram shown in the figure is that of a
If the bending moment diagram for a simply suppoted beam is of the form as given in figure then the load acting on the beam is
A beam is simply supported at its ends and is loaded by a couple at its mid-span as shown in figure. Shear force diagram for the beam is given by the figure
A horizontal beam carrying uniformly distributed load is supported with equal overhangs as shown in the figure beiow. The resultant bending moment at the mid-span shall be zero if a/b is
Since both the support reactions and loadings are symmetrically place about the center both the support will be of same magnitude and equal to ½ of total load acting on beam.
RA = RB = [w (2a+b)] / 2
= (w (2a+b))/2=w(a+b/2)
From the fig. bending moment at center (C) i.e. mid-span will be equal to:
(RA) (b/2) – w (a+b/2)( ((a+b/2))/2 ) = BMc =0
w(a+b/2) (b/2) – w (a+b/2)( ((a+b/2))/2 ) =0
b = 2a
a/b = 1/2