Unit 5 — Internal Resultants

Shear & Moment Equations
and Diagrams

Visualize how internal shear forces and bending moments vary along a beam under any combination of loads.

📌 Most Important Note

Answering Shear & Moment equations and diagrams questions accurately is mostly dependent on drawing the FBD correctly (and obtaining support reactions accurately) and being able to transform w(x) → V(x) → M(x) correctly using the differential and integral relations. Master these two skills and the rest follows.

⚡ Interactive Diagram — choose load type and adjust parameters

The FBD → w(x) → V(x) → M(x) Chain Master This

Every shear and moment problem follows the same transformation pipeline. Understanding each step makes the whole problem routine.

The Transformation Pipeline

The workflow for any beam problem is a strict sequence:

FBD of entire beam → find support reactions (Ay, By, MA…)
From reactions + loading → w(x) — identify the distributed load function on each segment
w(x) → V(x) via dV/dx = −w(x), or equivalently ΔV = −∫w dx
V(x) → M(x) via dM/dx = V(x), or equivalently ΔM = ∫V dx
Verify boundary conditions at supports and free ends

dV/dx = −w(x)   // slope of SFD = −(distributed load intensity)
dM/dx = V(x)    // slope of BMD = shear force at that point
ΔV = −∫w(x) dx // change in V = −area under load diagram
ΔM = ∫V(x) dx   // change in M = area under shear diagram

Sign Convention Critical — Never Skip

All results are meaningless without a consistent, stated sign convention.

✅ Positive Shear V

Upward force on the left face of the cut; downward on the right face. Tends to cause clockwise rotation of the free body element.

✅ Positive Moment M (Sagging)

Concave upward bending — the beam "smiles." Compresses top fibers. This is positive in the standard beam convention.

Load Type → Diagram Shape Reference

w(x) Loading	V(x) Shape (SFD)	M(x) Shape (BMD)
None (w=0)	Constant (flat)	Linear (sloped line)
Uniform (UDL, w=const)	Linear (1st degree)	Parabolic (2nd degree)
Triangular (w∝x)	Parabolic (2nd degree)	Cubic (3rd degree)
Concentrated force P	Jump of magnitude P	Kink (slope change)
Concentrated moment M₀	No change	Jump of magnitude M₀

Worked Example 1 Simply Supported + Point Load

Full FBD → w(x) → V(x) → M(x) pipeline for a simply supported beam with a single concentrated force.

Worked Example 1Simply Supported Beam — Concentrated Load P = 60 kN at x = 4 m

GIVEN: L = 10 m | P = 60 kN downward | a = 4 m from A

FBD of entire beam — find reactions

Apply ΣM_B = 0 (moments about B to eliminate By):

Ay × 10 = 60 × (10−4) = 360 → Ay = 36 kN ↑
ΣFy = 0: By = 60 − 36 = 24 kN ↑

Identify w(x) — no distributed load on either segment

Segment 0 < x < 4: w(x) = 0 // no distributed load
Segment 4 < x < 10: w(x) = 0 // no distributed load

w(x) → V(x): constant in each segment, jump at point load

V(0⁺) = +Ay = +36 kN            // starts at reaction
V(4⁻) = +36 kN                   // no change (w=0)
V(4⁺) = 36 − 60 = −24 kN     // jump down by P
V(10) = −24 + 24 = 0 ✓

V(x) → M(x): area under SFD accumulates moment

M(0) = 0 // BC: pin end
M(4) = 0 + 36×4 = +144 kN·m // area of +36 rectangle
M(10) = 144 + (−24)×6 = 0 ✓ // BC: roller end

Complete beam + SFD + BMD stacked summary

Worked Example 2 Cantilever + UDL

Full pipeline for a cantilever beam with a uniform distributed load — w(x)→V(x)→M(x) with integration.

Worked Example 2Cantilever Beam — UDL w₀ = 8 kN/m over full span

GIVEN: Fixed at A (left), free at B (right) | L = 6 m | w₀ = 8 kN/m downward

FBD — fixed wall provides vertical force Ay and moment MA

W = w₀ × L = 8 × 6 = 48 kN (resultant at midspan x=3m)
ΣFy=0: Ay = 48 kN ↑
ΣM_A=0: MA = w₀L²/2 = 8×36/2 = 144 kN·m ↺

w(x) = 8 kN/m constant — uniform load over full span

w(x) = 8 kN/m // constant, downward, 0 ≤ x ≤ 6 m
dV/dx = −w(x) = −8 // slope of SFD is −8 everywhere

w(x) → V(x): V(x) = 48 − 8x (linear, slope = −8)

V(x) = Ay − w·x = 48 − 8x
V(0) = +48 kN, V(3) = +24 kN, V(6) = 0 ✓

V(x) → M(x): integrate linear V → parabolic M

M(x) = MA + ∫₀ˣ V dx = 144 + 48x − 4x²
M(0) = 144 kN·m (fixed-end), M(6) = 144+288−144 = 0 ✓ (free end)

⚠ Note: M is maximum at the fixed end A, not mid-span. For a cantilever, the wall must resist the full accumulated moment from all the distributed load.

Worked Example 3 Triangular Load → V(x) → M(x)

Non-uniform w(x) produces increasingly complex V and M — showing the full chain with integration.

Worked Example 3Simply Supported — Triangular Load (0 at A, w₀ = 12 kN/m at B)

GIVEN: L = 9 m | w(x) = (4/3)x kN/m | zero at A, 12 kN/m at B

FBD — resultant of triangular load acts at 2L/3 from A

w(x) = (w₀/L)·x = (12/9)x = (4/3)x kN/m
Resultant W = ½·w₀·L = ½×12×9 = 54 kN at x = 2L/3 = 6 m from A
ΣM_A=0: By×9 = 54×6 = 324 → By = 36 kN
ΣFy=0: Ay = 54−36 = 18 kN

w(x) = (4/3)x — linearly increasing load

w(x) = (4/3)x kN/m // linear, 0 at x=0, 12 at x=9
dV/dx = −w(x) = −(4/3)x // slope of V is also linear → V is parabolic

w(x) → V(x) = 18 − (2/3)x² — parabolic SFD

V(x) = Ay − ∫₀ˣ w dx = 18 − (2/3)x²
V(0)=+18 kN, V(9)=18−54=−36+36=0 ✓
V=0 at x=√27 = 3√3 ≈ 5.20 m ← location of M_max

V(x) → M(x) = 18x − (2/9)x³ — cubic BMD

M(x) = ∫₀ˣ V dx = 18x − (2/9)x³
M(0)=0, M(9)=162−162=0 ✓
M_max at x=5.20m: M=18(5.20)−(2/9)(5.20)³ = 93.6−32.2 = ≈ 62.4 kN·m

Key insight: triangular load (linear w) → parabolic V → cubic M. Each integration raises the polynomial degree by 1.

Boundary Conditions Summary

Pin / Roller End

V = reaction
M = 0

Free End

V = 0
M = 0

Fixed End

V = reaction shear
M = reaction moment

Practice Problems

Seven problems covering FBD, w(x)→V(x)→M(x) transformations, and visual diagram identification.

Score

0 / 7

Problem 01 — FBD & Reactions

Simply Supported Beam — Find Reactions

A simply supported beam of length L = 12 m carries a concentrated load P = 72 kN at x = 3 m from the left support A. Draw the FBD and find A_y.

ΣM_B=0: Ay×12 = 72×(12−3) = 648 → Ay = 54 kN. Then By = 72−54 = 18 kN.

Problem 02 — w(x) → V(x)

Shear Force from UDL Using the Differential Relation

A simply supported beam (L=10 m) carries a uniform load w = 5 kN/m over its full length. Reactions: Ay = By = 25 kN. Using V(x) = Ay − w·x, find V at x = 3 m.

V(x) = Ay − w·x // from integrating dV/dx = −w with V(0) = Ay

V(3) = 25 − 5×3 = 25 − 15 = 10 kN.

Problem 03 — V(x) → M(x) Area Method

Bending Moment via Area Under SFD

Continuing from P2 (UDL w=5 kN/m, L=10 m, Ay=25 kN). The shear function is V(x) = 25−5x (linear). Using ΔM = ∫V dx, find M at x = 5 m (mid-span).

M(5) = ∫₀⁵ V dx = ∫₀⁵ (25−5x) dx = [25x − 2.5x²]₀⁵

kN·m

M(5) = 25(5) − 2.5(25) = 125 − 62.5 = 62.5 kN·m. Also = wL²/8 = 5×100/8 = 62.5 ✓

Problem 04 — FBD + Full Pipeline

Cantilever — Fixed-End Moment from w(x)→V(x)→M(x)

A cantilever beam is fixed at A (left), free at B (right). Length L = 5 m, uniform load w = 10 kN/m. Using the pipeline, find the magnitude of M at x = 0 (fixed end).

From free end: V(5)=0, V(0)=w×L=10×5=50kN
M(5)=0, M(0) = M(5)−∫₀⁵V dx

kN·m

Working from free end B: V(x from B) = w·s where s is distance from B. M at fixed end = ½·w·L² = ½×10×25 = 125 kN·m. Or: ∫₀⁵(50−10x)dx = [50x−5x²]₀⁵ = 250−125=125.

Problem 05 — Find Location of Maximum Moment

Position Where V(x) = 0 (Triangular Load)

A simply supported beam (L=9m) has a triangular load: w(x) = (4/3)x kN/m (zero at A, 12 kN/m at B). Reactions: Ay=18 kN, By=36 kN. The shear is V(x) = 18 − (2/3)x². Find the value of x (in m) where V(x) = 0 (maximum moment location). Round to 2 decimal places.

Set V(x) = 0: 18 − (2/3)x² = 0 → x² = 27 → x = ?

18 = (2/3)x² → x² = 27 → x = √27 = 3√3 ≈ 5.196 ≈ 5.20 m

Problem 06 — Visual: Identify the Correct SFD

Which SFD matches this beam?

A simply supported beam carries a uniform distributed load w over its full span. The reactions are equal (Ay = By = wL/2). Which of the four SFDs below is correct?

Option A

Option B

Option C

Option D

Problem 07 — Visual: Identify the Correct BMD

Which BMD matches a simply supported beam with a single point load at mid-span?

The beam has a concentrated load P at the center (x = L/2). SFD has a jump at center. Which BMD is correct?

Option A

Option B

Option C

Option D

Assessment — Unit 5

Shear & Moment Diagrams

12 questions · Mixed types including visual · 20 minute limit · 70% to pass

Questions

20:00

Time Limit

70%

To Pass

5× Multiple Choice 4× Numeric 1× True/False 2× Visual

Exam in Progress

12 questions — FBD, transformation pipeline, diagrams. Lock-in on Submit.

20:00

Question 01 of 12

Multiple Choice

What does the differential relation dV/dx = −w(x) tell us about the Shear Force Diagram?

Question 02 of 12

Numeric

A simply supported beam (L = 10 m) carries P = 80 kN at x = 2 m from A. Perform the FBD and find A_y (in kN). ΣM_B = 0.

Question 03 of 12

Multiple Choice

A beam segment has no distributed load (w = 0). What is the shape of the Shear Force Diagram in that segment?

Question 04 of 12

Numeric

A simply supported beam (L=8m) carries a UDL of w=6 kN/m over the full span. Reactions: Ay=By=24 kN. Using V(x) = Ay − w·x, find V at x = 6 m (in kN). Include sign.

Question 05 of 12

Multiple Choice

In the area method, the change in bending moment ΔM between two points equals:

Question 06 of 12

Numeric

A beam segment has a constant shear V = +40 kN over a length of 3 m. The moment at the left end of this segment is M₁ = 50 kN·m. Using the area method (ΔM = V·Δx), find M₂ at the right end (in kN·m).

kN·m

Question 07 of 12

True / False

A concentrated moment M₀ applied at a point on a beam causes a step (jump) in the Bending Moment Diagram but does NOT cause a jump in the Shear Force Diagram.

Question 08 of 12

Numeric

A cantilever beam (fixed at left, free at right) has length L = 4 m and UDL w = 15 kN/m. Using the pipeline: Ay = w·L = 60 kN, M_A = w·L²/2. Find |M_A| (in kN·m).

kN·m

Question 09 of 12

Multiple Choice

A beam segment has a triangular distributed load (linearly increasing from 0 to w₀). The Shear Force Diagram in this segment is:

Question 10 of 12

Visual

A cantilever beam is fixed at the left end A and free at B. A uniform distributed load acts downward over the full span. Which diagram shows the correct shape of the Bending Moment Diagram? (Note: M=0 at free end, M=max at fixed end.)

Option A

Option B

Option C

Option D

Question 11 of 12

Multiple Choice

To find the location of maximum bending moment on a beam, you should look for the point where:

Question 12 of 12

Visual

A cantilever beam is fixed at left (A) and free at right (B). A single concentrated force P acts downward at the free end B. Which SFD is correct? (No distributed load exists — so V is constant throughout.)

Option A

Option B

Option C

Option D

Exam Complete

Shear & Moment Equationsand Diagrams

The Transformation Pipeline

✅ Positive Shear V

✅ Positive Moment M (Sagging)

Pin / Roller End

Free End

Fixed End

Simply Supported Beam — Find Reactions

Shear Force from UDL Using the Differential Relation

Bending Moment via Area Under SFD

Cantilever — Fixed-End Moment from w(x)→V(x)→M(x)

Position Where V(x) = 0 (Triangular Load)

Which SFD matches this beam?

Which BMD matches a simply supported beam with a single point load at mid-span?

Shear & Moment Diagrams

Exam in Progress

Shear & Moment Equations
and Diagrams