Concept Notes
1. Shear Force (S.F)
- Definition: The algebraic sum of all vertical forces acting either to the left or right of a section of a beam.
- Formula:
Sign Convention:
- Upward force on left side of section = Positive.
- Downward force on left side of section = Negative.
Example: A simply supported beam of span 6 m carries a point load of 30 kN at the center.
- Reactions = 15 kN each.
- At left support, S.F just left of load = +15 kN, just right = -15 kN.
2. Bending Moment (B.M)
- Definition: The algebraic sum of moments of all forces about a section of the beam.
- Formula:
Sign Convention:
- Sagging (concave up) → Positive B.M.
- Hogging (concave down) → Negative B.M.
Example: Same beam as above:
- Bending Moment at center =
3. Relation between Load, Shear Force, and Bending Moment
where www = intensity of distributed load.
4. Important Cases
- Point Load:
- S.F changes suddenly (step).
- B.M is linear.
- UDL (Uniformly Distributed Load):
- S.F is linear.
- B.M is quadratic (parabolic).
- UVL (Uniformly Varying Load):
- S.F is quadratic.
- B.M is cubic.
5. Key Diagrams
- Shear Force Diagram (SFD): Graphical representation of variation of shear force along beam.
- Bending Moment Diagram (BMD): Graphical representation of bending moment variation along beam.
👉 Drawing SFD & BMD is scoring in ECET because 1–2 problems are guaranteed.
⚙️ Formulas
For simply supported beam with central load WWW:
For simply supported beam with UDL www per unit length:
🔟 10 MCQs
Q1. Shear force at a section is:
a) Sum of horizontal forces
b) Sum of vertical forces
c) Sum of moments
d) None
Q2. Bending moment at a section is:
a) Sum of vertical forces
b) Sum of horizontal forces
c) Algebraic sum of moments about section
d) Product of force × area
Q3. For a simply supported beam with a central point load W, maximum B.M = ?
a) WL
b) WL/2
c) WL/4
d) WL/8
Q4. In a simply supported beam with span L and UDL w, maximum B.M = ?
a) 
b) 
c) 
d) 
Q5. If shear force at a section is zero, bending moment is:
a) Zero
b) Maximum or minimum
c) Uniform
d) Negative
Q6. Relation between S.F and load intensity is:
a) 
b)
c)
d) (b) and (c)
Q7. For a cantilever with end point load W, maximum B.M at fixed end = ?
a) WL
b) WL/2
c) WL/4
d) WL/8
Q8. In SFD of a point load, variation is:
a) Linear
b) Constant (step)
c) Parabolic
d) Cubic
Q9. In BMD of UDL, variation is:
a) Linear
b) Step
c) Parabolic
d) Cubic
Q10. In sign convention, sagging moment is taken as:
a) Positive
b) Negative
c) Zero
d) None
✅ Answer Key
| Q | Answer |
|---|---|
| 1 | b |
| 2 | c |
| 3 | c |
| 4 | b |
| 5 | b |
| 6 | d |
| 7 | a |
| 8 | b |
| 9 | c |
| 10 | a |
🧠 Explanations
- Q1: Shear force = sum of vertical forces → (b).
- Q2: B.M = algebraic sum of moments → (c).
- Q3: Max BM in central load = WL/4 → (c).
- Q4: For UDL, max BM = wL²/8 → (b).
- Q5: Where SF = 0, BM is extreme (max/min) → (b).
- Q6: Relations are: dV/dx = -w and dM/dx = V → (d).
- Q7: Cantilever max BM = WL at fixed end → (a).
- Q8: Point load causes sudden change in SF (step) → (b).
- Q9: For UDL, BM diagram is parabolic → (c).
- Q10: Sagging = Positive → (a).
🎯 Motivation / Why Practice Matters
SFD and BMD problems are 100% guaranteed in ECET Civil.
- They check your ability to apply equilibrium equations and draw diagrams quickly.
- Many students confuse sign convention, so practicing daily makes it automatic.
👉 If you can master basics, you’ll never lose marks in these direct formula-based problems.
📲 CTA
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