Some Guidelines for Constructing Shear Force and Bending Moment Diagrams - Sign Conventions and Boundary Conditions

Oct 29, 2026

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Guidelines explain how to set up sign conventions and boundary conditions for shear force (V) and bending moment (M) diagrams of a beam. The material lists reactions and relationships between shear, moment, deflection, and slope at pinned and roller supports, as well as at free ends. A worked example demonstrates finding support reactions, drawing the shear force diagram step by step, then constructing the bending moment diagram using the area under the shear curve, verifying end conditions.

LanguageEnglish

CategoryResearch & Report

FormatPDF

Pages3

Reading Time8 min

AccessFree to download

Sign Conventions

Boundary Conditions

Worked Example: Reaction Forces

Construction of the Shear Force Diagram

Construction of the Bending Moment Diagram

Some Guidelines for Constructing Shear Force and Bending Moment Diagrams SIGN CONVENTIONS + M + M -M -V -M BOUNDARY CONDITIONS A B Ax Bx A B Ax Bx Ay By | Pinned LEFT End:
Reaction force, Ax is unknown.
Reaction force, Ay is unknown. V (shear force) = Ay (reaction force) M = 0 (unless there is an applied moment at this point) Deflection, y = 0
Slope, 􀁔 is unknown | Roller Support at Right End:
Reaction force, Bx =0 .
Reaction force, By is unknown. V (shear force) = -By (reaction force) M = 0 (unless there is an applied moment at this point) Deflection, y = 0
Slope, 􀁔 is unknown | | --- | --- | | | | MA Ax A B Free End: V (shear force) = 0 (unless there is an applied point force at this end) M = 0 (unless there is an applied moment at this point) Deflection, y is unknown Slope, 􀁔 is unknown Example: Draw the shear force and bending moment diagram for the following beam: Step 1: Find the reaction forces at A and D. Draw F.B.D.: 􀂦 Fx = 0, Ax = 0 􀂦 MA = 0, (20 kips)(6 ft) + (12 kips)(14 ft) + (1.5 kips/ft)(8 ft)(28ft)–(Dy)(24 ft) = 0 Dy = 26 kips 􀂦 Fy = 0, Ay + Dy – 20 kips – 12 kips –(1.5 kips/ft)(8 ft) = 0, Ay = 18 kips Construction of the shear force diagram 20 kips 12 kips 1.5 kips/ft V 18 kips | | | 12 kips | | x | | --- | --- | --- | --- | --- | | | | | | | | -2 kips | | | | | | | -14 kips | | | | | | | | | | a) V = Ay = 18 kips at point A. b) The shear force is constant until there is another load applied at B. c) The shear force decreases by 20 kips to –2 kips at B because of the applied 20 kip force in the negative y direction. d) The shear force is constant until there is another load applied at C. e) The shear force decreases by 12 kips to –14 kips at C because of the applied 12 kip force in the negative y direction. f) The shear force is constant until there is another load applied at D. g) The shear force increases by 26 kips to 12 kips at D because of the 26 kip reaction force in the positive y direction. h) The shear force decreases linearly from D to E because there is a constant applied load in the negative y-direction. i) The change in shear force from D to E is equal to the area under the load curve between D and E, -12 kips,[ADE = (-1.5 kips/ft)(8 ft) = -12 kips] j) The shear force at E = 0 as expected by inspection of the boundary conditions. Construction of the Bending Moment diagram V 18 kips | B
6 ft
| | 8 ft | C | 10 ft

12 kips | E 8 ft | x
| | --- | --- | --- | --- | --- | --- | --- | | | | | | |
| | | | | | | | | | | -2 kips | | | | | | | | | -14 | | kips | | | | | | | | | | | | M |
108 kip-ft | | | | x
| | --- | --- | --- | --- | --- | | |
92 kip | -ft | | | | | | | | | | | | -48 kip-ft | | | a) M = 0 at point A because it is a pinned end with no applied bending moment. b) MB = Ma + (the area under the shear force diagram between A and B.) c) MB = 0 + (18 kips)(6 ft) = 108 kip-ft d) MC = MB + (the area under the shear force diagram between B and C.) e) MC = 108 kip-ft- (2 kips)(8 ft) = 92 kip-ft f) MD = MC + (the area under the shear force diagram between C and D.) g) MD = 92 kip-ft- (14 kips)(10 ft) = -48 kip-ft h) ME = MD + (the area under the shear force diagram between D and E.) i) ME = -48 kip-ft + 1/2 (12 kips)(8 ft) = 0 kip-ft (as expected)

FAQs

What shear and moment values apply at a pinned left end?

At the pinned left end, reaction forces Ax and Ay are treated as unknown, with V equal to Ay and M equal to 0 unless an applied moment exists at that point. Deflection y is set to 0 and the slope is unknown.

How are boundary conditions handled for a roller support at the right end?

For a roller support at the right end, Bx equals 0 while By is unknown. The relationship uses V equal to -By, and M equals 0 unless an applied moment exists. Deflection y is set to 0 and the slope is unknown.

How do you construct the bending moment diagram from the shear force diagram?

Compute bending moment changes by adding the area under the shear force diagram over each segment. The worked example uses MB = Ma + area(A to B) and continues similarly through C, D, and E, then checks that M matches the expected end conditions (e.g., M = 0 at E).

Oct 29, 2026

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