Login
OR
Don't worry. We'll fix this for you!
In civil engineering, beams are fundamental structural elements designed to support loads by resisting bending. Imagine a wooden plank laid across two supports; when weight is placed on it, the plank bends slightly but holds the load. This simple example illustrates the basic function of a beam.
Beams are everywhere: in bridges, buildings, and even furniture. Understanding how beams behave under various loads is crucial for designing safe and economical structures. This section covers the types of beams, the kinds of loads they carry, how to find the reactions at supports, and how to analyze internal forces such as shear force and bending moment. We will also learn to draw their corresponding diagrams, which are essential tools in structural design.
Beams differ mainly by how they are supported or fixed at their ends. These boundary conditions affect how the beam carries loads and the internal forces developed.
Simply Supported Beam: Supported at both ends, usually with a pin support on one side and a roller on the other. The beam is free to rotate and has no moment resistance at supports. Common in bridges and simple floor spans.
Cantilever Beam: Fixed at one end and free at the other. The fixed end resists rotation and moment. Used in balconies, overhangs, and some bridge sections.
Fixed Beam: Both ends are fixed, preventing rotation and translation. This creates moments at supports and reduces deflections. Used where rigidity is required, such as in continuous beams over multiple supports.
Loads on beams can vary in type and distribution. Understanding these helps predict how the beam will react internally.
Point Load: A load applied at a single point on the beam, such as a heavy machine resting on a floor beam.
Uniformly Distributed Load (UDL): Load spread evenly across a length of the beam, like the weight of a floor slab or a row of books on a shelf. It is expressed in force per unit length (e.g., N/m).
Varying Load: Load intensity changes along the length, often triangular or trapezoidal, such as wind pressure on a sloped roof.
To analyze a beam, the first step is to find the support reactions-the forces and moments the supports exert on the beam to keep it in equilibrium.
We use the principles of static equilibrium, which state that for a body at rest:
These conditions ensure the beam does not move or rotate.
graph TD A[Draw Free Body Diagram] --> B[Identify Supports and Loads] B --> C[Apply \u03a3F_x = 0] C --> D[Apply \u03a3F_y = 0] D --> E[Apply \u03a3M = 0] E --> F[Calculate Support Reactions]
A Free Body Diagram (FBD) is a sketch showing the beam isolated from its surroundings, with all applied loads and unknown support reactions indicated. This is the foundation for applying equilibrium equations.
When a beam carries loads, internal forces develop within its cross-section to resist bending and shear. Two primary internal forces are:
Sign Conventions: For consistency, the following sign rules are commonly used:
Correct use of sign conventions is essential to avoid errors in diagrams and calculations.
Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) graphically represent how shear force and bending moment vary along the length of the beam. These diagrams help engineers identify critical points where maximum stresses occur.
To construct these diagrams:
The SFD shows a sudden drop at the point load, indicating a change in shear force. The BMD is a smooth curve, peaking at the center where bending moment is maximum.
Step 1: Draw the free body diagram showing the beam with supports at A and B, and the point load P = 12 kN at 3 m from each support.
Step 2: Apply equilibrium equations.
Sum of vertical forces: \(\sum F_y = 0\)
\(R_A + R_B - 12 = 0\) (1)
Sum of moments about A: \(\sum M_A = 0\)
\(R_B \times 6 - 12 \times 3 = 0\)
\(6 R_B = 36 \Rightarrow R_B = 6 \text{ kN}\)
From (1): \(R_A + 6 = 12 \Rightarrow R_A = 6 \text{ kN}\)
Answer: Reactions at supports are \(R_A = 6 \text{ kN}\), \(R_B = 6 \text{ kN}\).
Step 1: Calculate support reactions.
Total load \(W = w \times L = 4 \times 8 = 32 \text{ kN}\).
Since the load is symmetrical, reactions are equal:
\(R_A = R_B = \frac{32}{2} = 16 \text{ kN}\).
Step 2: Calculate shear force at key points.
At A (x=0): \(V = +16 \text{ kN}\).
At distance \(x\) from A: Shear force decreases by load intensity:
\(V = 16 - 4x\).
At B (x=8 m): \(V = 16 - 4 \times 8 = 16 - 32 = -16 \text{ kN}\).
Step 3: Calculate bending moment at key points.
At A and B: \(M = 0\) (simply supported).
At distance \(x\):
\(M = R_A \times x - w \times \frac{x^2}{2} = 16x - 2x^2\).
Maximum moment occurs at \(x = \frac{R_A}{w} = \frac{16}{4} = 4 \text{ m}\).
Maximum moment:
\(M_{\max} = 16 \times 4 - 2 \times 4^2 = 64 - 32 = 32 \text{ kNm}\).
Answer: Support reactions are 16 kN each. Shear force decreases linearly from +16 kN at A to -16 kN at B. Maximum bending moment of 32 kNm occurs at mid-span (4 m).
Step 1: Draw the free body diagram showing the fixed support at A and the load at free end B.
Step 2: Calculate support reactions at fixed end A.
Vertical reaction \(R_A = 10 \text{ kN}\) upwards to balance load.
Moment reaction \(M_A = 10 \times 4 = 40 \text{ kNm}\) counterclockwise to balance moment caused by load.
Step 3: Shear force and bending moment at any section \(x\) from fixed end:
Shear force \(V = -10 \text{ kN}\) constant along length (negative sign indicates direction).
Bending moment \(M = -10 \times x\), maximum at fixed end \(x=4\) m is \(-40 \text{ kNm}\).
Answer: Support reaction at fixed end is 10 kN upward and moment reaction is 40 kNm counterclockwise. Shear force is constant at 10 kN along the beam, and bending moment varies linearly from 0 at free end to 40 kNm at fixed end.
Step 1: Calculate total load.
\(W = w \times L = 3 \times 6 = 18 \text{ kN}\).
Step 2: Fixed end moments for uniform load are given by formulas:
\[ M_A = M_B = -\frac{w L^2}{12} = -\frac{3 \times 6^2}{12} = -9 \text{ kNm} \]
Negative sign indicates hogging moment (moment causing compression at bottom fibers).
Step 3: Maximum bending moment occurs at a distance \(x = \frac{L}{2} = 3 \text{ m}\) and is given by:
\[ M_{\max} = \frac{w L^2}{24} = \frac{3 \times 6^2}{24} = 4.5 \text{ kNm} \]
This is a sagging moment (positive moment causing compression at top fibers).
Answer: Fixed end moments are \(-9 \text{ kNm}\) at both supports, and maximum positive bending moment of \(4.5 \text{ kNm}\) occurs at mid-span.
Step 1: Calculate total UDL load.
\(W_{UDL} = 2 \times 10 = 20 \text{ kN}\).
Step 2: Draw free body diagram with point load at 4 m and UDL over 10 m.
Step 3: Let reactions at supports be \(R_A\) and \(R_B\).
Sum of vertical forces:
\(R_A + R_B = 20 + 20 = 40 \text{ kN}\) (1)
Sum of moments about A:
\(R_B \times 10 - 20 \times 4 - 20 \times 5 = 0\)
\(10 R_B = 80 + 100 = 180\)
\(R_B = 18 \text{ kN}\)
From (1): \(R_A = 40 - 18 = 22 \text{ kN}\)
Step 4: Calculate shear force at key points:
Step 5: Calculate bending moment at key points:
\(M = R_A \times 4 - 2 \times \frac{4^2}{2} = 22 \times 4 - 2 \times 8 = 88 - 16 = 72 \text{ kNm}\)
\(M = 22 \times 5 - 20 \times 1 - 2 \times \frac{5^2}{2} = 110 - 20 - 25 = 65 \text{ kNm}\)
Answer: Reactions are \(R_A = 22 \text{ kN}\), \(R_B = 18 \text{ kN}\). Shear force diagram shows a drop at the point load, and bending moment diagram peaks near the point load location.
When to use: At the beginning of any beam analysis problem
When to use: While drawing shear force and bending moment diagrams
When to use: When calculating reactions or internal forces for uniformly distributed loads
When to use: During quick calculations in entrance exams
When to use: After computing support reactions
Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
Go to practice →
Your Score
Your Rank
| Rank | Name | Score |
|---|
