Bending Moment and Shear Force Diagrams

TL;DR

Bending Moment and Shear Force Diagrams (BMD and SFD) visually represent the internal forces within a beam under load. They are essential for understanding how a beam resists external forces and for designing safe structures. You'll learn to calculate shear force and bending moment at different points and then plot these values along the beam's length.

1. The Mental Model

Imagine a beam as a long, thin stick you're trying to break. A Shear Force Diagram shows you where you're trying to cut it, and a Bending Moment Diagram shows you where you're trying to bend it.

2. The Core Material

When a beam is subjected to external loads (like weights or pressures), internal forces develop within its cross-section to resist these loads. These internal forces are primarily shear force (V) and bending moment (M).

• Shear Force (V): This is the internal transverse force acting parallel to the beam's cross-section, tending to shear (cut) the beam. Think of it as the force trying to push one part of the beam up and the other part down.
• Bending Moment (M): This is the internal couple (moment) acting about the neutral axis of the beam's cross-section, tending to bend the beam. Think of it as the force trying to make the beam curve.

The diagrams are simply plots of these internal forces along the length of the beam.

Sign Conventions

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Consistent sign conventions are crucial for drawing correct diagrams. We'll use the following:

• Shear Force: Positive when the left-hand part of the beam tends to move upwards relative to the right-hand part. Alternatively, an upward external force on the left or a downward external force on the right is positive shear.
• Bending Moment: Positive when the beam sags (concave up), meaning tension at the bottom and compression at the top (often called "sagging" moment). Negative when the beam hogs (concave down), meaning compression at the bottom and tension at the top ("hogging" moment).

Steps to Draw SFD and BMD

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1. Calculate Support Reactions: First, treat the beam as a rigid body and use static equilibrium equations ($\Sigma F_y = 0$, $\Sigma M = 0$) to find the unknown support reactions.
2. Section the Beam: Divide the beam into segments based on changes in loading or geometry (e.g., concentrated loads, distributed loads, supports).
3. Cut and Analyze: For each segment, make an imaginary "cut" at a distance 'x' from one end (usually the left). Consider the forces to the left (or right) of the cut.
4. Derive Equations:
   • Shear Force (V(x)): Sum all vertical forces to the left (or right) of the cut, applying your sign convention.
   • Bending Moment (M(x)): Sum all moments about the cut point due to forces to the left (or right) of the cut, applying your sign convention.
5. Plot the Diagrams:
   • Plot V(x) vs. x for the SFD.
   • Plot M(x) vs. x for the BMD.

Relationships between Load, Shear Force, and Bending Moment

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These relationships provide a quick check for your diagrams:

• The slope of the SFD at any point is equal to the negative of the distributed load intensity at that point: $dV/dx = -w(x)$.
• The slope of the BMD at any point is equal to the shear force at that point: $dM/dx = V(x)$.
• A concentrated load causes a sudden jump (discontinuity) in the SFD equal to the load's magnitude.
• A concentrated moment causes a sudden jump in the BMD equal to the moment's magnitude.
• The change in shear force between two points is the negative of the area under the distributed load curve between those points.
• The change in bending moment between two points is the area under the shear force diagram between those points.

Here's a flowchart of the process:

graph TD
    A["Identify Beam, Loads, and Supports"] --> B["Calculate Support Reactions (Equilibrium)"]
    B --> C{"Any Changes in Loading / Supports?"}
    C -- Yes --> D["Divide Beam into Segments"]
    C -- No --> E["Treat as Single Segment"]
    D --> F["For Each Segment: Make an Imaginary Cut"]
    E --> F
    F --> G["Derive Shear Force Equation V(x)"]
    F --> H["Derive Bending Moment Equation M(x)"]
    G --> I["Plot V(x) vs. x (SFD)"]
    H --> J["Plot M(x) vs. x (BMD)"]
    I --> K["Review & Check Diagrams (Slopes, Jumps, Areas)"]
    J --> K

3. Worked Example

Let's consider a simply supported beam of length $L=6m$ with a concentrated load $P=10kN$ at its mid-span ($x=3m$).

1. Support Reactions:

   • Let $R_A$ be the reaction at the left support (pin) and $R_B$ at the right support (roller).
   • $\Sigma F_y = 0 \implies R_A + R_B - P = 0$
   • $\Sigma M_A = 0 \implies P \times (L/2) - R_B \times L = 0 \implies 10 \times 3 - R_B \times 6 = 0 \implies 30 - 6R_B = 0 \implies R_B = 5 kN$.
   • Substituting $R_B$ back: $R_A + 5 - 10 = 0 \implies R_A = 5 kN$.

2. Section the Beam:

   • Segment 1: $0 \le x < 3m$ (from A to P)
   • Segment 2: $3m < x \le 6m$ (from P to B)

3. Derive Equations:

   • Segment 1 ($0 \le x < 3m$):

     • Shear Force V(x): Looking left, only $R_A$ is acting upwards. So, $V(x) = +R_A = +5 kN$. (Constant)
     • Bending Moment M(x): Looking left, $M(x) = R_A \times x = +5x \ kN \cdot m$. (Linear, starts at 0, goes to $5 \times 3 = 15 kN \cdot m$ at $x=3m$)

   • Segment 2 ($3m < x \le 6m$):

     • Shear Force V(x): Looking left, $V(x) = +R_A - P = +5 - 10 = -5 kN$. (Constant)
     • Bending Moment M(x): Looking left, $M(x) = R_A \times x - P \times (x - 3) = 5x - 10(x - 3) = 5x - 10x + 30 = -5x + 30 \ kN \cdot m$. (Linear, at $x=3m$, $M = -5(3)+30 = 15 kN \cdot m$. At $x=6m$, $M = -5(6)+30 = 0 kN \cdot m$)

4. Plot the Diagrams:

   • SFD:

     • Starts at $+5kN$ (due to $R_A$).
     • Constant at $+5kN$ until $x=3m$.
     • Drops by $10kN$ (due to $P$) at $x=3m$, from $+5kN$ to $-5kN$.
     • Constant at $-5kN$ until $x=6m$.
     • Jumps up by $5kN$ (due to $R_B$) at $x=6m$, from $-5kN$ to $0$.

   • BMD:

     • Starts at $0$ at $x=0$.
     • Increases linearly from $0$ to $+15kN \cdot m$ at $x=3m$.
     • Decreases linearly from $+15kN \cdot m$ to $0$ at $x=6m$.
     • The maximum bending moment is $+15kN \cdot m$ at mid-span.

4. Key Takeaways

• SFD and BMD are graphical representations of internal shear force and bending moment along a beam.
• Always start by calculating the support reactions using equilibrium equations.
• Consistently apply a sign convention for shear and bending moment.
• The slope of the BMD is the shear force, and the slope of the SFD is the negative of the distributed load.
• Concentrated loads cause sudden "jumps" in the SFD, and concentrated moments cause "jumps" in the BMD.
• The bending moment is zero at simply supported or free ends unless a concentrated moment is applied there.

Common Mistakes to Avoid

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• Incorrect Sign Conventions: Be consistent! Swapping conventions mid-problem leads to inverted diagrams.
• Arithmetic Errors: Double-check your basic algebra, especially when calculating reactions and moments.
• Ignoring Units: Always include units for all values and your final diagrams.
• Drawing Smooth Curves Instead of Lines: For concentrated loads/moments or uniform distributed loads, sections of the SFD and BMD should be straight lines or parabolas, not arbitrary curves.
• Not Closing to Zero: Both SFD and BMD should ideally close to zero at the ends of the beam, assuming it's in equilibrium. If they don't, you've likely made a calculation error.

5. Now Try It

For a cantilever beam fixed at one end (say, at $x=0$) and free at the other ($x=L=4m$), subjected to a uniformly distributed load of $w = 2 kN/m$ along its entire length.

1. Calculate the fixed-end reactions (vertical force and moment).
2. Derive the shear force $V(x)$ and bending moment $M(x)$ equations along the beam.
3. Sketch the SFD and BMD.

What success looks like: Your SFD should be a linear slope from $wL$ at the fixed end to $0$ at the free end. Your BMD should be a parabolic curve, starting at a maximum negative value (hogging) at the fixed end and going to $0$ at the free end.