The maximum bending moment in a simply supported beam length L loaded by a concentrated load W at midpoint is given by
WL/4
This explanation focuses on finding the maximum bending moment experienced by a simply supported beam when subjected to a single concentrated load placed exactly at its center.
We consider a beam that is supported at both ends (simply supported) and has a total length denoted by L. A concentrated force, represented as W, is applied exactly at the midpoint of the beam's length (at distance L/2 from either support).
Due to the symmetrical loading (the load W is at the center), the total load W is distributed equally between the two supports. Let the reactions at the supports be RA and RB.
Using the principle of equilibrium (sum of vertical forces = 0):
$$ R_A + R_B = W $$
Since the load is at the midpoint, the reactions are equal:
$$ R_A = R_B $$
Substituting this into the equilibrium equation:
$$ R_A + R_A = W \implies 2R_A = W \implies R_A = \frac{W}{2} $$
Therefore, the reaction at each support is W/2.
The bending moment at any section of the beam is the algebraic sum of the moments of the forces to the left (or right) of that section. Let's consider a section at a distance x from the left support (where 0 ≤ x ≤ L/2).
The bending moment M(x) at this section is caused by the reaction force RA acting at a distance x:
$$ M(x) = R_A \times x $$
Substituting the value of RA:
$$ M(x) = \frac{W}{2} \times x $$
This formula gives the bending moment for any point between the left support and the center of the beam.
The bending moment varies linearly from 0 at the left support (x=0) to its maximum value at the center of the beam (x=L/2). To find the maximum bending moment (Mmax), we substitute x = L/2 into the moment equation:
$$ M_{max} = M\left(\frac{L}{2}\right) = \frac{W}{2} \times \frac{L}{2} $$
$$ M_{max} = \frac{WL}{4} $$
The maximum bending moment in a simply supported beam of length L, carrying a concentrated load W at its midpoint, occurs at the center and is equal to WL/4.
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