Shear Force & Bending Moment Diagrams
Step-by-Step Method for Simply Supported Beams, Cantilevers & Overhanging Beams — UDL, Point Loads, Couples & Solved Examples
Last Updated: March 2026
📌 Key Takeaways
- The Shear Force Diagram (SFD) shows how the internal shear force varies along the length of a beam. The Bending Moment Diagram (BMD) shows how the internal bending moment varies.
- Sign convention: Upward forces on the left of a section → positive shear. Sagging (concave up) bending → positive moment. Memorise this — inconsistent sign convention is the most common source of errors.
- Under a point load, the SFD has a sudden vertical jump equal to the load magnitude. The BMD has a kink (change in slope) at that point.
- Under a UDL, the SFD varies linearly and the BMD varies parabolically (2nd degree curve).
- The maximum bending moment occurs where the shear force is zero (or changes sign) — this is the most important relationship in beam analysis.
- The slope of the BMD at any point = shear force at that point: dM/dx = V. The slope of the SFD = intensity of distributed load: dV/dx = −w.
- SFD and BMD are essential inputs for beam design — the maximum BM determines the required section modulus, and the maximum SF determines the shear reinforcement.
1. Definitions — Shear Force & Bending Moment
When a beam is loaded, it develops internal forces at every cross-section to maintain equilibrium. To find these internal forces at any section, we use the method of sections — we imagine cutting the beam at that point and apply equilibrium to one of the two resulting free bodies. The two most important internal forces in a beam are shear force and bending moment.
Shear Force (V or S) — Definition
The shear force at any cross-section of a beam is the algebraic sum of all transverse forces acting on the beam to one side of that section.
It represents the tendency of the two parts of the beam (on either side of the cut) to slide vertically relative to each other.
Units: kN or N
The shear force has the same magnitude whether calculated from the left or right side — but must respect the sign convention to get the correct sign.
Bending Moment (M) — Definition
The bending moment at any cross-section of a beam is the algebraic sum of the moments of all forces acting on the beam to one side of that section, taken about the centroid of that section.
It represents the tendency of the beam to bend (rotate) at that section.
Units: kN·m or N·m
Like shear force, the bending moment gives the same magnitude from either side — the sign depends on the convention used.
The Shear Force Diagram (SFD) is a graph plotted along the length of the beam with the beam axis as the x-axis and the shear force value as the y-axis. Similarly, the Bending Moment Diagram (BMD) plots the bending moment value along the beam length. Together, these two diagrams give a complete picture of how the beam is stressed internally under the applied loading.
2. Sign Convention
The sign convention must be defined before any calculation. The most widely used convention in structural engineering — and the one used in all GATE CE problems — is as follows:
| Quantity | Positive (+) | Negative (−) |
|---|---|---|
| Shear Force | Forces on the left side of the section acting upward, OR forces on the right side acting downward — the section tends to rotate clockwise. | Forces on the left acting downward, OR forces on the right acting upward — the section tends to rotate anticlockwise. |
| Bending Moment | Sagging — beam concave upward (like a smile). Bottom fibre in tension, top fibre in compression. Produced when the left side has a net clockwise moment about the section. | Hogging — beam concave downward (like a frown). Top fibre in tension, bottom fibre in compression. Produced when the left side has a net anticlockwise moment about the section. |
Memory aid for shear force sign: “Left up = positive shear.” If all forces on the left of a section are upward (net), the shear is positive. This is consistent with the clockwise rotation tendency.
Memory aid for bending moment sign: “Sagging = positive, hogging = negative.” A simply supported beam with downward loads always sags (positive BM) between the supports. A cantilever with a downward load at the free end always hogs (negative BM).
Important: The sign convention is a matter of definition — what matters is consistency throughout a problem. Some textbooks use the opposite convention for shear force. Always check which convention is being used before starting a problem, and stick to it throughout.
3. Key Mathematical Relationships
Three fundamental differential relationships connect the distributed load intensity, shear force, and bending moment. These relationships are the key to understanding the shape of SFD and BMD without doing full calculations — and they are regularly tested in GATE CE as conceptual MCQ questions.
The Three Fundamental Relationships
Relationship 1 — Load and Shear Force:
dV/dx = −w(x)
The slope of the SFD at any point equals the negative of the distributed load intensity at that point. Where there is no distributed load (w = 0), the SFD is horizontal (constant shear). Where there is a UDL (w = constant), the SFD is a straight sloping line.
Relationship 2 — Shear Force and Bending Moment:
dM/dx = V
The slope of the BMD at any point equals the shear force at that point. Where V = 0, the BMD has a horizontal tangent — this is the location of maximum (or minimum) bending moment.
Relationship 3 — Load and Bending Moment (combined):
d²M/dx² = −w(x)
Under a UDL, the BMD is parabolic (2nd degree). Under no load, the BMD is linear. Under a linearly varying load, the BMD is cubic.
Integral Relationships — Area Under SFD = Change in BM
V₂ − V₁ = −∫w dx (change in SF = negative area under load diagram)
M₂ − M₁ = ∫V dx (change in BM = area under SFD between two points)
These relationships allow you to construct the BMD directly from the SFD by calculating areas — without writing moment expressions. This is the fastest method for GATE numerical problems.
| Loading Type | SFD Shape | BMD Shape |
|---|---|---|
| No load (free span) | Horizontal line (constant) | Straight inclined line (linear) |
| Point load | Sudden vertical jump at load point | Kink (change of slope) at load point |
| Uniformly Distributed Load (UDL) | Straight inclined line (linear) | Parabolic curve (2nd degree) |
| Uniformly Varying Load (UVL / triangular) | Parabolic curve (2nd degree) | Cubic curve (3rd degree) |
| Applied couple (moment) | No change (SFD unaffected) | Sudden vertical jump at couple location |
4. Step-by-Step Method for Drawing SFD & BMD
Follow this systematic procedure for every beam problem — whether in an exam or in practice. Skipping any step is the most common source of errors.
- Step 1 — Draw the free body diagram (FBD): Sketch the beam with all applied loads (magnitudes, types, positions) and mark the unknown support reactions (RA, RB, MA, etc.). Show the positive directions for all reactions.
- Step 2 — Calculate support reactions: Apply equilibrium equations (ΣFy = 0, ΣM = 0). For a simply supported beam, take moments about one support to find the other reaction, then use ΣFy = 0 for the first. Always verify by taking moments about the second support.
- Step 3 — Identify critical sections: Mark all points where the loading changes — supports, point load locations, start and end of UDLs, applied couple locations. These are the sections where you must calculate V and M values.
- Step 4 — Calculate shear force at each critical section: Starting from the left end, calculate V just to the left and just to the right of each critical section. Use the sign convention consistently: sum all upward forces to the left as positive.
- Step 5 — Plot the SFD: Plot the shear force values calculated above. Connect them with straight lines (under no load or point loads) or parabolic curves (under UDL). Label all key values including where V = 0.
- Step 6 — Calculate bending moment at each critical section: At each critical section, M = sum of moments of all forces to the left of the section (upward forces × distance − downward forces × distance). Alternatively, use M₂ = M₁ + area under SFD between the two sections.
- Step 7 — Find location and magnitude of maximum BM: The maximum BM occurs where V = 0 (or changes sign). If V = 0 within a UDL region, find the exact position by interpolation or by writing the expression for V(x) and setting it to zero.
- Step 8 — Plot the BMD: Plot the bending moment values. Connect with straight lines (under no distributed load), parabolic curves (under UDL), or cubic curves (under UVL). Plot sagging moments below the beam axis (positive convention) and hogging moments above.
- Step 9 — Check: The BM must be zero at any pin or roller support and at any free end. The fixing moment at a fixed support equals the BM value calculated there. Verify these boundary conditions before finalising.
5. Effect of Different Load Types on SFD & BMD Shape
Point Load
At the location of a concentrated point load P, the shear force diagram has an abrupt vertical jump of magnitude P (upward reaction → upward jump; downward load → downward jump). The bending moment diagram has a sharp kink (change in gradient) at that point — the slope of the BMD changes abruptly because V changes abruptly.
Uniformly Distributed Load (UDL)
Over a region with UDL of intensity w (kN/m), the shear force varies linearly (dV/dx = −w = constant). The bending moment varies parabolically (d²M/dx² = −w). The parabola opens downward for a downward UDL. The vertex of the parabola (point of maximum BM) occurs where V = 0.
Uniformly Varying Load (Triangular Load)
Over a region with a triangular load (zero at one end, maximum at the other), the shear force varies parabolically and the bending moment varies as a cubic curve. The resultant of a triangular load acts at one-third of the span from the larger end.
Applied Couple (Moment Load)
An externally applied couple M₀ at a point on the beam does not affect the shear force at that point — the SFD is unchanged and continuous at that location. However, the bending moment diagram has an abrupt vertical jump of magnitude M₀ at that point. The direction of jump (up or down) depends on the direction of the applied couple.
Summary — Load Type vs Diagram Shape
No load region → SFD: constant horizontal | BMD: linear (inclined)
Point load → SFD: vertical jump | BMD: kink (slope change)
UDL → SFD: linear | BMD: parabolic (2nd degree)
UVL (triangular) → SFD: parabolic | BMD: cubic (3rd degree)
Applied couple → SFD: no change | BMD: vertical jump
6. Worked Example 1 — Simply Supported Beam with Two Point Loads
Problem: A simply supported beam AB of span 10 m carries point loads of 20 kN at C (3 m from A) and 30 kN at D (7 m from A). Draw the SFD and BMD. Find the maximum bending moment and its location.
Step 1 & 2 — Support Reactions
ΣMA = 0: RB × 10 = (20 × 3) + (30 × 7) = 60 + 210 = 270
RB = 27 kN
ΣFy = 0: RA = 20 + 30 − 27 = 23 kN
Check (ΣMB = 0): 23 × 10 = 230 = (20 × 7) + (30 × 3) = 140 + 90 = 230 ✓
Step 3–5 — Shear Force Diagram
At A (just right of support): V = +23 kN (reaction RA = 23 kN upward)
Just left of C (x = 3⁻): V = +23 kN (no loads between A and C)
Just right of C (x = 3⁺): V = 23 − 20 = +3 kN (20 kN downward load at C)
Just left of D (x = 7⁻): V = +3 kN (no loads between C and D)
Just right of D (x = 7⁺): V = 3 − 30 = −27 kN (30 kN downward load at D)
At B (just left of support): V = −27 kN
At B (just right, including reaction): V = −27 + 27 = 0 ✓
SFD shape: Horizontal at +23 from A to C; drops to +3 at C; horizontal at +3 from C to D; drops to −27 at D; horizontal at −27 from D to B.
Step 6–8 — Bending Moment Diagram
M at A = 0 (pin support — moment is zero)
M at C = RA × 3 = 23 × 3 = +69 kN·m (sagging)
M at D = RA × 7 − 20 × 4 = 161 − 80 = +81 kN·m (sagging)
Alternatively using area under SFD: MD = MC + (3 × +3) = 69 + 9 = +78 kN·m…
Let us recalculate directly: M at D = 23×7 − 20×4 = 161 − 80 = 81 kN·m
M at B = 0 (roller support — moment is zero)
Check from right: M at D = RB × 3 = 27 × 3 = 81 kN·m ✓
Maximum BM = 81 kN·m at D (x = 7 m from A)
Note: V changes sign at D (from +3 to −27) — confirming that maximum BM occurs at D.
BMD shape: Straight line from 0 at A to 69 at C; straight line from 69 at C to 81 at D; straight line from 81 at D back to 0 at B.
7. Worked Example 2 — Simply Supported Beam with Full-Span UDL
Problem: A simply supported beam AB of span 6 m carries a UDL of 12 kN/m over the entire span. Draw the SFD and BMD. Find the maximum shear force and maximum bending moment.
Solution
Reactions (by symmetry): RA = RB = (12 × 6)/2 = 36 kN
SFD — Shear force at distance x from A:
V(x) = RA − w·x = 36 − 12x
V at A (x=0) = +36 kN | V at midspan (x=3) = 36 − 36 = 0 | V at B (x=6) = 36 − 72 = −36 kN
Maximum shear force = 36 kN (at supports A and B)
SFD shape: Straight line from +36 at A, decreasing linearly to 0 at midspan, then to −36 at B.
BMD — Bending moment at distance x from A:
M(x) = RA·x − w·x²/2 = 36x − 6x²
M at A (x=0) = 0 | M at x=1: 36(1) − 6(1) = 30 kN·m | M at x=2: 72 − 24 = 48 kN·m
M at x=3 (midspan): 36(3) − 6(9) = 108 − 54 = 54 kN·m
Check using formula: Mmax = wL²/8 = 12×36/8 = 432/8 = 54 kN·m ✓
M at B (x=6) = 36(6) − 6(36) = 216 − 216 = 0 ✓
BMD shape: Parabolic curve (opening downward) from 0 at A, rising to maximum of 54 kN·m at midspan, back to 0 at B.
Location of maximum BM: Where V = 0 → 36 − 12x = 0 → x = 3 m (midspan, as expected for symmetric loading).
8. Worked Example 3 — Simply Supported Beam with UDL + Point Load
Problem: A simply supported beam AB of span 8 m carries a UDL of 10 kN/m over the entire span and a central point load of 40 kN at midspan (C, 4 m from A). Draw SFD and BMD. Find Mmax.
Solution
Reactions (by symmetry):
Total UDL = 10 × 8 = 80 kN | Point load = 40 kN | Total = 120 kN
RA = RB = 120/2 = 60 kN
SFD:
V at A (just right) = +60 kN
V at C (just left, x = 4⁻) = 60 − 10(4) = 60 − 40 = +20 kN
V at C (just right, x = 4⁺) = 20 − 40 = −20 kN (40 kN point load drops the SFD)
V at B (just left) = −60 + 10(0) = −60 kN… recalculating:
V at B (just left) = 60 − 10(8) − 40 + 60 from right… using left side: V = 60 − 10(8) = −20…
Correct approach from left at x = 8⁻: V = 60 − 10(8) − 40 = 60 − 80 − 40 = −60 kN
V at B (including RB) = −60 + 60 = 0 ✓
V = 0 occurs at x = 4 m (midspan, by symmetry) — confirmed by SFD changing sign at C.
BMD:
M(x) for 0 ≤ x ≤ 4: M = 60x − 10x²/2 = 60x − 5x²
M at A = 0 | M at C (x=4): 60(4) − 5(16) = 240 − 80 = 160 kN·m
Check using superposition: Mmax = WL/4 + wL²/8 = (40×8)/4 + (10×64)/8 = 80 + 80 = 160 kN·m ✓
BMD shape: Parabolic from A to C (due to UDL), with a kink at C (due to point load), then mirror-image parabola from C to B.
9. Worked Example 4 — Cantilever Beam with UDL and Point Load
Problem: A cantilever beam AC is fixed at A and free at C. Total span = 6 m. A point load of 15 kN acts at midspan B (3 m from A), and a UDL of 8 kN/m acts from B to C (over the outer 3 m). Draw SFD and BMD.
Solution — Working from the Free End (C)
For cantilevers, it is easier to work from the free end (C) towards the fixed end (A) — reactions at A are not needed initially.
Reactions at A:
RA = 15 + (8 × 3) = 15 + 24 = 39 kN (upward)
MA = 15 × 3 + 8 × 3 × (3 + 1.5) = 45 + 8 × 3 × 4.5 = 45 + 108 = 153 kN·m (hogging)
SFD (working left to right):
V at A (just right) = +39 kN (reaction upward)
Between A and B: No load → V = constant = +39 kN
At B (just right): V = 39 − 15 = +24 kN
V at any point x between B and C (measuring from A): V = 39 − 15 − 8(x − 3) = 24 − 8(x−3)
V at C (x = 6): 24 − 8(3) = 24 − 24 = 0 ✓ (no force at free end)
SFD shape: Constant +39 from A to B; drops to +24 at B; decreases linearly to 0 at C.
BMD (working from free end C):
M at C = 0 (free end)
M at B (from right side) = −8 × 3 × 1.5 = −36 kN·m (hogging)
M at A (from right side) = −15 × 3 − 8 × 3 × 4.5 = −45 − 108 = −153 kN·m (hogging)
BMD shape: Parabolic from C (zero) to B (−36 kN·m); then linear from B to A (−153 kN·m).
The entire BMD is negative (hogging) — characteristic of a cantilever under downward loads.
10. Worked Example 5 — Overhanging Beam
Problem: Beam ABCD has supports at A (pin) and C (roller). Spans: AB = 2 m, BC = 4 m, CD = 2 m (overhang). Loads: 20 kN downward at B, UDL of 5 kN/m over BC. Draw SFD and BMD. Identify the point of contraflexure.
Solution
Reactions:
ΣMA = 0: RC × 6 = 20 × 2 + 5 × 4 × 4 = 40 + 80 = 120 → RC = 20 kN
ΣFy = 0: RA = 20 + 5×4 − 20 = 20 kN → RA = 20 kN
SFD (left to right):
V just right of A = +20 kN
V just left of B = +20 kN
V just right of B = 20 − 20 = 0 kN
V at any point x in BC (measuring from A, 2 ≤ x ≤ 6): V = 20 − 20 − 5(x−2) = −5(x−2)
V at C (x = 6): −5(4) = −20 kN (just left of C)
V just right of C = −20 + 20 = 0 kN (no load on CD overhang)
V at D = 0 (free end, no load on overhang)
BMD:
M at A = 0 | M at B = 20 × 2 = +40 kN·m (sagging)
M at x in BC: M = 20x − 20(x−2) − 5(x−2)²/2 = 40 − 5(x−2)²/2
M at C (x=6): 40 − 5(16)/2 = 40 − 40 = 0 kN·m ✓ (roller support)
M at D = 0 (free end)
Maximum BM: V = 0 just right of B (x = 2 m) → Mmax = +40 kN·m at B
Point of contraflexure: M = 0 within BC: 40 − 5(x−2)²/2 = 0 → (x−2)² = 16 → x−2 = 4 → x = 6 m (at C)
In this case, BM is positive (sagging) throughout BC and zero only at the supports — no internal contraflexure point within the span BC.
11. Worked Example 6 — Beam with Applied Couple
Problem: A simply supported beam AB of span 8 m has an applied clockwise couple of 40 kN·m at C, located 3 m from A. No other loads except the support reactions. Draw the SFD and BMD.
Solution
Reactions:
ΣMA = 0: RB × 8 = 40 (clockwise couple at C acts clockwise on beam)
RB = 5 kN (upward)
ΣFy = 0: RA = −5 kN → RA = 5 kN downward
(The couple alone causes the beam to want to rotate, so reactions are equal and opposite — a couple themselves.)
SFD:
The applied couple does not cause any transverse force, so the SFD is unaffected by the couple itself.
V just right of A = −5 kN (RA is downward, so shear is negative from left)
V is constant at −5 kN throughout (no distributed loads)
V just right of B = −5 + 5 = 0 ✓
BMD:
M at A = 0
M just left of C (x = 3⁻): M = −5 × 3 = −15 kN·m (hogging)
M just right of C (x = 3⁺): M = −15 + 40 = +25 kN·m (sagging) — the clockwise couple causes a sudden upward jump of 40 kN·m in the BMD.
M at B (x = 8): +25 − 5×5 = 25 − 25 = 0 ✓
BMD shape: Linear from 0 at A to −15 kN·m just left of C; jumps to +25 kN·m just right of C; linear back to 0 at B.
Point of contraflexure: Between A and C (BM changes from negative to positive at C). The BMD crosses zero at x = 3 m (at C, the jump location).
12. Common Mistakes Students Make
- Inconsistent sign convention: The most frequent error. Switching between “left up positive” and “right up positive” mid-problem leads to wrong signs throughout. Define your sign convention at the start and apply it uniformly to every section. Write it at the top of your solution as a reminder.
- Not finding support reactions before drawing the SFD: Many students attempt to draw the SFD for a simply supported beam without first calculating RA and RB. The starting value of the SFD at A depends on RA — without it, nothing that follows is correct.
- Drawing a straight BMD under a UDL: Under a uniformly distributed load, the BMD is parabolic — not a straight line. A straight BMD under UDL is always wrong. The parabola’s vertex (maximum or minimum) is where V = 0.
- Forgetting that the SFD is unaffected by an applied couple: An applied moment (couple) causes a sudden jump in the BMD but has absolutely no effect on the SFD. Many students mistakenly show a jump in the SFD at the couple location.
- Confusing the location of maximum BM: Maximum BM does not always occur at midspan. It occurs where V = 0 (or changes sign). For asymmetric loading, this is almost never at midspan. Always find where V = 0 first, then calculate M there.
13. Frequently Asked Questions
Why does maximum bending moment occur where shear force is zero?
This follows directly from the mathematical relationship dM/dx = V. At the location of maximum bending moment, the slope of the BMD is zero — meaning the BMD has a horizontal tangent at that point. Since dM/dx = V, a zero slope of the BMD means V = 0. This is exactly analogous to finding the maximum of any function by setting its derivative to zero. The practical implication is powerful: you do not need to write a full expression for M(x) — just find where V = 0 in the SFD, and that is where maximum bending moment occurs.
What is a point of contraflexure and how do you find it?
A point of contraflexure is a location along the beam where the bending moment is zero and changes sign — from sagging (positive) to hogging (negative) or vice versa. To find it, write the bending moment expression M(x) for the relevant section, set M(x) = 0, and solve for x. Verify that the sign of M changes at this point (not just touches zero). In simply supported beams under downward loads, the BM is always positive between the supports — there is no contraflexure point. Contraflexure points appear in overhanging beams, continuous beams, and fixed beams.
How does the SFD and BMD change if a support settles (sinks)?
For a statically determinate beam (simply supported, cantilever, overhanging), support settlement has no effect on the SFD and BMD. This is because the reactions in a determinate structure are found entirely from equilibrium, which does not involve the actual positions of the supports. For a statically indeterminate beam (fixed, propped cantilever, continuous), support settlement changes the reactions and therefore completely alters the SFD and BMD. This is one of the key practical disadvantages of highly indeterminate structures built on compressible soils.
Can the shear force be zero throughout a beam?
Yes — if only an applied couple (moment load) acts on a simply supported beam with no transverse loads, the reactions are a couple (equal and opposite vertical forces), and the shear force is constant and non-zero throughout. However, if no transverse loads act at all (including no reactions), then V = 0 everywhere. A more interesting case: a free-floating beam with only two equal and opposite couples applied has V = 0 everywhere but non-zero bending moments between the couple locations. In practice, V = 0 throughout a real loaded beam is rare.