Shear force and bending moment diagrams

Why it matters

Before you can size a beam, a vehicle chassis rail, an axle or a leaf spring, you need to know where the internal shear force and bending moment are largest and what their values are. Shear force and bending moment diagrams (SFD and BMD) give exactly that, and their shapes also tell you where the beam sags, where it hogs and where it changes curvature. Every later beam topic, bending stress, shear stress and deflection, starts from these diagrams.

Key ideas

Internal forces. Cut a loaded beam at a section x. To keep either piece in equilibrium, the cut face must carry a shear force V (transverse) and a bending moment M. They are found from the free-body diagram of either side; both sides must give the same answer.

Sign convention (the common one in Indian textbooks).

• Shear force is positive when the resultant of forces to the left of the section acts upward (left side up, right side down).
• Bending moment is positive (sagging) when it makes the beam concave upward, i.e. compression at the top. Hogging moments are negative. Any convention works if used consistently; GATE answers usually just need magnitude and location.

Load, shear and moment relations. With w the downward distributed load intensity:

• dV/dx = −w: the slope of the SFD equals minus the load intensity.
• dM/dx = V: the slope of the BMD equals the shear force. Consequences:
• Under no load, V is constant and M varies linearly.
• Under a UDL, V varies linearly and M is a parabola (second degree).
• Under a linearly varying load, V is parabolic and M is cubic.
• A point load causes a jump in the SFD equal to the load; the BMD has a kink (change of slope) there.
• A concentrated couple causes a jump in the BMD equal to the couple, with no effect on the SFD at that point.
• Bending moment is maximum or minimum where the shear force is zero or changes sign.
• The change in M between two sections equals the area of the SFD between them.

Point of contraflexure. A point where the bending moment changes sign (sagging to hogging). It occurs in overhanging, propped and fixed beams. It is a good place for a joint because the moment there is zero.

Standard results worth memorising.

• Simply supported, central point load W: M_max = W·L/4 at mid-span.
• Simply supported, point load W at distance a from one end and b from the other: M_max = W·a·b/L under the load.
• Simply supported, UDL w over the span: M_max = w·L²/8 at mid-span; V_max = w·L/2 at supports.
• Cantilever, end load W: M_max = W·L at the fixed end.
• Cantilever, UDL w: M_max = w·L²/2 at the fixed end.

Procedure. Find reactions; mark every point where the loading changes (point loads, start or end of distributed loads, supports, couples); find V and M just to the left and right of each; join them with the shape the load dictates; find zero-shear points for maxima and zero-moment points for contraflexure.

Formulas

dV/dx = −w, dM/dx = V, d²M/dx² = −w

• V: shear force (N); M: bending moment (N·m); w: distributed load, downward positive (N/m); x: position along the beam (m).

M₂ − M₁ = ∫ V dx (area of the SFD between sections 1 and 2)

M_max = W·L/4, M_max = W·a·b/L, M_max = w·L²/8 (simply supported) M_max = W·L, M_max = w·L²/2 (cantilever, at the fixed end)

• W: point load (N); L: span (m); a, b: distances of the load from the supports (m).

x₀ = R_A / w

• Position of zero shear from the left support for a simply supported beam with a UDL starting at A (and no other load before x₀); then M_max = R_A² / (2·w).

Worked examples

Example 1 (standard): point load plus UDL. Given: simply supported beam AB, span 6 m, UDL 5 kN/m over the whole span and a 20 kN point load at C, 2 m from A. Draw the SFD and BMD and find the maximum bending moment.

1. Reactions, moments about A: R_B·6 = 20·2 + (5·6)·3 = 40 + 90 = 130, so R_B = 21.67 kN; R_A = 50 − 21.67 = 28.33 kN.
2. Shear: at A, V = +28.33 kN. Just left of C: 28.33 − 5·2 = 18.33 kN. Just right of C: 18.33 − 20 = −1.67 kN. At B (left side): −1.67 − 5·4 = −21.67 kN, which equals −R_B (check).
3. The shear changes sign at C, so M_max is at C.
4. M_C = R_A·2 − 5·2·1 = 56.67 − 10 = 46.67 kN·m. Answer: M_max = 46.7 kN·m at the point load; SFD linear between ±21.67 kN with a 20 kN drop at C; BMD parabolic segments, zero at both supports.

Example 2 (GATE level): overhanging beam with contraflexure. Given: beam ABC, supports at A and B with AB = 6 m, overhang BC = 2 m. UDL 10 kN/m on AB only, and a 20 kN point load at the free end C. Find the reactions, maximum sagging and hogging moments and the point of contraflexure.

1. Moments about A: R_B·6 = 60·3 + 20·8 = 180 + 160 = 340, so R_B = 56.67 kN; R_A = 80 − 56.67 = 23.33 kN.
2. Shear in AB: V = 23.33 − 10·x, zero at x₀ = 2.333 m.
3. Maximum sagging moment: M = R_A² / (2w) = 23.33² / 20 = 27.22 kN·m at 2.333 m from A.
4. Moment at B (from the overhang side): M_B = −20 × 2 = −40 kN·m (hogging).
5. Shear checks: just left of B, 23.33 − 60 = −36.67 kN; just right of B, −36.67 + 56.67 = +20 kN, constant to C, then drops to zero at the 20 kN load.
6. Contraflexure: M(x) = 23.33·x − 5·x² = 0 gives x = 4.667 m from A. Answer: R_A = 23.33 kN, R_B = 56.67 kN; +27.2 kN·m at 2.33 m from A; −40 kN·m at B (the design moment); contraflexure 4.67 m from A.

Common mistakes

• Taking the maximum moment where the load is applied without checking where V = 0.
• Forgetting the jump in the BMD caused by an applied couple.
• Replacing a UDL by its resultant and then using that point load when computing the SFD inside the loaded length.
• Mixing sign conventions between the left and right free bodies.
• Assuming the maximum bending moment of an overhanging beam is the sagging one; the hogging moment over the support is often larger.
• Reading the SFD slope as the load magnitude but with the wrong sign.

For GATE ME

Expect maximum bending moment or its location for combinations of point loads, UDLs and couples; identifying a loading from a given SFD or BMD shape; points of contraflexure in overhanging beams; and the effect of an applied couple. Practise using dM/dx = V and area-of-SFD checks to avoid sign slips.

Quick check

1. A 4 m simply supported beam carries a 2 kN/m UDL. Find M_max.
2. What does a vertical jump in the BMD indicate?
3. A cantilever of length 3 m carries 15 kN at its free end. Find the maximum moment.
4. Under a UDL, what shape is the BMD?

Answers: 1. 2 × 16/8 = 4 kN·m. 2. A concentrated couple applied at that point. 3. 45 kN·m at the fixed end (hogging). 4. Parabolic.