7-80 Draw the Shear and Moment Diagrams for the Beam: A Step‑by‑Step Guide
To draw the shear and moment diagrams for the beam identified as 7‑80, you must first understand the loading configuration, support conditions, and the mathematical relationships that govern internal forces. This guide walks you through every stage—from selecting the appropriate coordinate system to plotting the final diagrams—ensuring that the resulting sketches are both accurate and easy to interpret.
Introduction to Shear and Moment Diagrams
Shear force and bending moment diagrams are essential tools in structural analysis. They provide a visual representation of how internal forces vary along the length of a beam, enabling engineers to identify critical sections for design and reinforcement. The 7‑80 beam typically refers to a simply supported beam with a point load of 7 kN applied 80 mm from the left support, but the methodology described here applies to any similar problem involving known loads and span lengths.
Understanding the Beam and Its Loading
Support Conditions
- Simply supported: The beam rests on two pins or rollers, allowing rotation but restricting vertical translation.
- Reactions: The supports generate vertical reaction forces that balance the external loads.
Load Description - Point load: A concentrated force of 7 kN acts downward at a distance of 80 mm (0.08 m) from the left support. - Span length: The total distance between the two supports is often denoted as L; for the 7‑80 case, L may be 1.2 m, but the exact value must be confirmed from the problem statement.
Geometry and Coordinate System
- Choose the origin at the left support.
- Measure the position x along the beam from the origin toward the right support.
- Keep the units consistent (e.g., meters for distances, kilonewtons for forces).
Step‑by‑Step Procedure
1. Calculate Support Reactions
Using equilibrium equations: - ΣF_y = 0 → R_A + R_B – 7 kN = 0
- ΣM_A = 0 → R_B·L – 7 kN·0.08 m = 0
Solve for R_B, then substitute back to find R_A. ### 2. Define the Shear Force Equation
- For 0 ≤ x < 0.08 m, the shear force V(x) equals the left reaction R_A.
- For 0.08 m ≤ x ≤ L, V(x) equals R_A minus the point load (7 kN).
3. Integrate to Obtain the Bending Moment Equation
- The bending moment M(x) is the integral of V(x) with respect to x.
- Apply the appropriate piecewise expressions for each region of the beam.
4. Determine Key Values at Critical Points
- Shear force at the supports, at the load application point, and at the far support.
- Bending moment at the same locations, especially the maximum moment which usually occurs under the point load.
5. Plot the Diagrams
- Shear diagram: Draw a horizontal line at V = R_A from x = 0 to x = 0.08 m, then a step down to V = R_A – 7 kN until x = L.
- Moment diagram: Plot a linear variation of M(x) in each region; the slope of the moment diagram corresponds to the shear value at that segment. ## Example: Drawing the Diagrams for a 7‑80 Beam
Assume a simply supported beam with a span of 1.2 m (1200 mm) and a 7 kN point load located 80 mm from the left support.
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Reactions:
- R_B = (7 kN × 0.08 m) / 1.2 m = 0.467 kN - R_A = 7 kN – 0.467 kN = 6.533 kN
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Shear Force:
- 0 ≤ x < 0.08 m → V = 6.533 kN (constant)
- 0.08 m ≤ x ≤ 1.2 m → V = 6.533 kN – 7 kN = –0.467 kN (constant)
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Bending Moment:
- For 0 ≤ x < 0.08 m: M(x) = 6.533 kN·x
- For 0.08 m ≤ x ≤ 1.2 m: M(x) = 6.533 kN·x – 7 kN·(x – 0.08 m)
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**Key Moments
