Test: Shear Force and Bending Moment - Civil Engineering (CE) MCQ

BM = P × x where P = force applied and x = distance of force application from point where BM is to be calculated

Calculation:
Load acted from centroid of the load triangle.
Load = WL/2 acted at L/3 distance from fixed end
 (as per sign convention -ve sign is used)

At the end supports, the moment (couple) developed is zero, because there is no distance to take the perpendicular acting load. As the distance is zero, the moment is obviously zero.

is the equation of parabola.

The point loads are applied symmetrically to the simply supported beam with respect to mid-span (or point C), so the vertical reaction at both supports is equal and is half the applied load.
V_A = V_B = P
Calculating the bending moment at point D

Calculating the bending moment at point C

The bending moment at points A and B is zero because supports A and B are simply supports (roller and hinge).

Hence the bending moment is constant in the middle one-third part of the beam.

In cantilever beams, the maximum shear force occurs at the fixed end. In the free end, there is zero shear force. As we need to convert the udl in to load, we multiply the length of the cantilever beam with udl acting upon. For maximum shear force to obtain we ought to multiply load and distance and it surely occurs at the fixed end (w×l).

For vertical equilibrium
R₁ + R₂ = P
Taking moment about R₁, we have


The bending moment at the point of application of load = Area under OABC

Bending moment Diagram:
Bending moment diagrams are simply plots of the bending moment (on the y-axis) versus the position of various points along the beam (on the x-axis).
For UDL(Uniformly Distributed Load)

Relationship between Shear force, Bending moment and Loading rate

• −(ds/dx) = w i.e. negative slope of Shear force diagram at any section will be equal to the load intensity at that section
• (dM/dx) = S i.e. The slope of Bending moment diagram at any section of a loaded beam will be equal to Intensity of shear force at that section

Additional Information
For point load

• Shear force is constant. So SFD is rectangular. Hence slope of SFD at any section will be 0.
• Bending Moment is of one degree. So BMD is Triangular and Slope of BMD at any section will be constant and equal to Shear force.

For UDL

• Shear force is of one degree. So SFD is Triangular and its slope at any section will be constant and equal to load intensity.
• Bending moment is of two degree. So BMD is parabolic and its slope at any section will be equal to shear force intensity at that section.

The positive bending moment is considered when it causes convexity downward or concavity at top. This is sagging. In simply supported beams, it occurs at mid span because the bending moment at the supports obviously will be zero hence the positive bending moment occurs in the mid span.

Here make hit and trial method
Consider first option (A) of moment diagram.

V = dM/dx
At right end B, slope must be zero as there is no shear force at B so option A is wrong.
Now, consider option (B)
Due to symmetric load intensity at left side too the shear force equal to zero so slope at left side must be zero. So option (B) wrong.

Now consider option (C)
Here at both ends, slope is zero means shear force is zero and also when we move from right to left, the rate of increase of shear force decreases due triangular shape of load intensity and at middle slope should be maximum and there after decreases so in option (C) all these criterion fulfills. Here (C) is the correct option.

As far as (D) option is concerned its middle part slope and right most part slope is strictly not agreeing with the load shown.

At support B, the BM is zero. The beam undergoes maximum BM at fixed end.
By joining the base line, free end and maximum BM point. We obtain a right angled triangle.