Edunes Online Education
B.Tech Mechanical Engg.
Engineering Mechanics : SFD and BMD
Cantilever Beam
Edunes Online Education
SHEAR FORCE & BENDING MOMENT – CANTILEVER BEAM
A Cantilever Beam is fixed at one end and free at the other.
At the fixed end, three reactions exist:
- Vertical Reaction (RA)
- Horizontal Reaction (HA)
- Reaction Moment (MR)
🧠 MEMORY IMAGE:
Think of your arm stretched out holding a weight.
Your shoulder = Fixed End.
Your palm = Free End.
The shoulder must resist force AND rotation.
STEP 1: HOW TO THINK (NOT JUST SOLVE)
Before writing equations, ask:
- Where is the fixed end?
- What forces exist on the left of my section?
- Is load point, UDL, or UVL?
- What is positive?
(Upward + , Downward −)
CORE IDEA:
Shear Force at section = Algebraic sum of vertical forces on LEFT side.
Bending Moment at section = Sum of (Force × distance) on LEFT side + applied moments.
Bending Moment at section = Sum of (Force × distance) on LEFT side + applied moments.
🧠 Golden Rule:
“CUT – ISOLATE – SUM – WRITE”
1. Cut the beam
2. Look LEFT
3. Sum forces
4. Write equation
“CUT – ISOLATE – SUM – WRITE”
1. Cut the beam
2. Look LEFT
3. Sum forces
4. Write equation
Draw shear force & bending moment diagrams and equations
Cantilever reactions and Momemts
SECTION AB (0 ≤ X ≤ 2)
For 0 ≤ X ≤ 2
Shear Force:
SF = RA = 130 kN
Bending Moment:
BM = − MR + RAX
BM = −720 + 130X
Shear Force:
SF = RA = 130 kN
Bending Moment:
BM = − MR + RAX
BM = −720 + 130X
| X | Shear Force (kN) | Bending Moment (kN·m) |
|---|---|---|
| 0 | 130 | −720 |
| 2 | 130 | −460 |
🧠 Observation:
If Shear is CONSTANT → SFD is straight line.
Moment is LINEAR → BMD is sloping line.
If Shear is CONSTANT → SFD is straight line.
Moment is LINEAR → BMD is sloping line.
SECTION BD (2 ≤ X ≤ 6)
Here UDL starts.
UDL effect = Load × length considered.
Shear Force:
SF = 130 − 20(X − 2)
Bending Moment:
BM = −720 + 130X − {20(X − 2)²}/2
SF = 130 − 20(X − 2)
Bending Moment:
BM = −720 + 130X − {20(X − 2)²}/2
| X | Shear Force (kN) | Bending Moment (kN·m) |
|---|---|---|
| 2 | 130 | −460 |
| 6 | 50 | −100 |
🧠 Observation:
UDL makes Shear vary LINEARLY.
UDL makes Moment PARABOLIC.
UDL makes Moment PARABOLIC.
SECTION DE (6 ≤ X ≤ 8)
Shear Force:
SF = 130 − 80 = 50 kN
Bending Moment:
BM = −720 + 130X − 80(X − 4)
SF = 130 − 80 = 50 kN
Bending Moment:
BM = −720 + 130X − 80(X − 4)
| X | Shear Force (kN) | Bending Moment (kN·m) |
|---|---|---|
| 6 | 50 | −100 |
| 8 | 50 | 0 |
🧠 IMPORTANT:
Moment at FREE END of a cantilever = ZERO (if no applied moment).
Moment at FREE END of a cantilever = ZERO (if no applied moment).
FINAL VISUAL LOGIC
Think Graphically:
• Point Load → Jump in SFD
• UDL → Sloping SFD
• Shear constant → Straight BMD
• Shear linear → Parabolic BMD
• Maximum BM occurs where SF = 0
• Point Load → Jump in SFD
• UDL → Sloping SFD
• Shear constant → Straight BMD
• Shear linear → Parabolic BMD
• Maximum BM occurs where SF = 0
🧠 Neurological Anchor:
Shear = “Force Feeling”
Moment = “Rotation Feeling”
If force is constant → moment changes smoothly.
If force changes → curvature appears.
Your brain remembers shapes better than numbers.
Shear = “Force Feeling”
Moment = “Rotation Feeling”
If force is constant → moment changes smoothly.
If force changes → curvature appears.
Your brain remembers shapes better than numbers.
EXAM THINKING STRATEGY
1. Draw FBD first.
2. Write reactions.
3. Divide into segments.
4. Write SF equation.
5. Integrate logic to get BM.
6. Verify boundary conditions.
2. Write reactions.
3. Divide into segments.
4. Write SF equation.
5. Integrate logic to get BM.
6. Verify boundary conditions.
🧠 Final Compression Formula:
Cantilever Fixed End → Maximum Moment
Free End → Zero Moment
UDL → Parabola
Point Load → Straight Line
Visualize before calculating.
Cantilever Fixed End → Maximum Moment
Free End → Zero Moment
UDL → Parabola
Point Load → Straight Line
Visualize before calculating.
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