A uniform beam of span l is rigidly fixed at both supports. It carries a uniformly distributed load w per unit length. The bending moment at mid-span is
wl2 / 24
Understanding Bending Moment in a Fixed Beam with Uniform Load
This question asks for the bending moment at the mid-span of a specific type of beam: a uniform beam that is rigidly fixed at both supports and subjected to a uniformly distributed load (UDL).
A beam fixed at both ends is known as a fixed-fixed beam or a propped cantilever if one end is fixed and the other simply supported. In this case, both ends are fixed, making it a fixed-fixed beam. The load is a uniformly distributed load, meaning the load is spread evenly across the entire span of the beam, with an intensity of 'w' per unit length.
Analyzing the Fixed-Fixed Beam Configuration
For a fixed beam carrying a uniformly distributed load 'w' over its entire span 'l', the support reactions and support moments are standard values derived from structural analysis principles (like consistent deformation or slope-deflection method). Due to symmetry in loading and support conditions, the vertical reactions at both fixed supports are equal, and the fixing moments at both supports are also equal.
- Total downward load = \(w \times l\)
- Vertical reactions at each support (\(R_A\), \(R_B\)) = \(\frac{wl}{2}\)
- Fixed end moments at each support (\(M_A\), \(M_B\)) = \(\frac{wl^2}{12}\) (These moments are typically considered negative or 'hogging' moments).
Calculating Bending Moment at Mid-Span
To find the bending moment at the mid-span, we can take a section at \(x = l/2\) from one of the supports, say support A. The bending moment at any section \(x\) from support A can be calculated by considering the forces and moments to the left of the section.
The components contributing to the bending moment at section \(x\) are:
- The fixed support moment at A (\(M_A\)).
- The vertical reaction at A (\(R_A\)).
- The load distributed over the length \(x\) from A (\(w \times x\)).
The bending moment \(M(x)\) at a distance \(x\) from support A is given by:
$$ M(x) = R_A \cdot x - M_A - (\text{Total UDL left of x}) \cdot (\text{Distance of UDL centroid from x}) $$
The total UDL left of \(x\) is \(w \cdot x\), and its centroid is at \(x/2\) from support A, or \(x - x/2 = x/2\) from the section at \(x\). Considering positive bending moment as sagging (smiley face shape) and negative as hogging (frowning face shape), and support moments typically being hogging (negative), the formula becomes:
$$ M(x) = R_A \cdot x - M_A - \frac{(w \cdot x) \cdot x}{2} $$
Substituting the values for \(R_A\) and \(M_A\):
$$ M(x) = \left(\frac{wl}{2}\right) \cdot x - \left(\frac{wl^2}{12}\right) - \frac{wx^2}{2} $$
We need the bending moment at mid-span, which is at \(x = l/2\). Substitute \(x = l/2\) into the equation:
$$ M\left(\frac{l}{2}\right) = \left(\frac{wl}{2}\right) \cdot \left(\frac{l}{2}\right) - \left(\frac{wl^2}{12}\right) - \frac{w\left(\frac{l}{2}\right)^2}{2} $$
$$ M_{mid-span} = \frac{wl^2}{4} - \frac{wl^2}{12} - \frac{w\left(\frac{l^2}{4}\right)}{2} $$
$$ M_{mid-span} = \frac{wl^2}{4} - \frac{wl^2}{12} - \frac{wl^2}{8} $$
To combine these terms, find a common denominator, which is 24:
$$ M_{mid-span} = \frac{6wl^2}{24} - \frac{2wl^2}{24} - \frac{3wl^2}{24} $$
$$ M_{mid-span} = \frac{(6 - 2 - 3)wl^2}{24} $$
$$ M_{mid-span} = \frac{1 \cdot wl^2}{24} $$
$$ M_{mid-span} = \frac{wl^2}{24} $$
This is the bending moment at the mid-span of the fixed-fixed beam under a uniformly distributed load.
Summary of Bending Moments
For a fixed beam with UDL 'w' over span 'l':
- Bending moment at supports (Fixed Ends) = \(-\frac{wl^2}{12}\) (Hogging)
- Bending moment at mid-span = \(+\frac{wl^2}{24}\) (Sagging)
Revision Table: Beam Bending Moments
| Beam Type | Loading | Maximum Bending Moment | Bending Moment at Mid-span |
|---|---|---|---|
| Simply Supported | Point Load (P) at mid-span | \(\frac{Pl}{4}\) (at mid-span) | \(\frac{Pl}{4}\) |
| Simply Supported | UDL (w) over full span | \(\frac{wl^2}{8}\) (at mid-span) | \(\frac{wl^2}{8}\) |
| Cantilever | Point Load (P) at free end | \(-Pl\) (at fixed end) | \(-P \cdot \frac{l}{2} = -\frac{Pl}{2}\) |
| Cantilever | UDL (w) over full span | \(-\frac{wl^2}{2}\) (at fixed end) | \(-\frac{w(l/2)^2}{2} = -\frac{wl^2}{8}\) |
| Fixed-Fixed | Point Load (P) at mid-span | \(\frac{Pl}{8}\) (at supports and mid-span, absolute value) | \(\frac{Pl}{8}\) (Sagging) |
| Fixed-Fixed | UDL (w) over full span | \(\frac{wl^2}{12}\) (at supports, absolute value) | \(\frac{wl^2}{24}\) (Sagging) |
Additional Information on Fixed Beams
Fixed beams are statically indeterminate structures. This means that the equations of static equilibrium alone (\(\sum F_y = 0\), \(\sum M = 0\)) are not sufficient to determine all the support reactions (vertical reactions and fixing moments). Additional equations based on the deformation of the beam are required. Common methods used to analyze indeterminate beams include:
- Slope-Deflection Method
- Moment Distribution Method
- Flexibility Method (Castigliano's Theorem, Unit Load Method)
- Stiffness Method
The formulas for reactions and moments in standard cases like a fixed beam with UDL are often memorized or found in structural analysis handbooks, as they are derived using these advanced methods. The presence of fixed supports significantly alters the bending moment and shear force diagrams compared to simply supported beams, typically resulting in smaller mid-span bending moments but introducing significant moments at the supports.
Important Questions from Bending Moment
Slope and deflection of a cantilever beam carrying a moment M at the free end is given by:
Which of the following beams is likely to have the point of contraflexure?
The point of contraflexure is the point at which ___________ changes its sign.
The maximum bending moment of the center of laminated spring of span L due to load W is given by-
If a simply supported beam is loaded with point load W at the centre then what is the ratio of bending moment at the support to the bending moment at the centre?
