# Shear and Moment Diagram Examples: 5 Beams Worked by Hand Source: https://webstructural.com/shear-and-moment-diagram-examples.html > Five shear and moment diagram examples worked step by step: two point loads, a cantilever, an overhanging beam, a triangular load and an applied moment, with reactions, diagrams and the largest moment. Five beams worked by hand, each showing one thing that trips people up: point loads, a cantilever, an overhang, a triangular load and an applied moment. Each one finds the reactions, draws the shear diagram, then the moment diagram, and finds the largest moment and where it is. You can build any of them in the [shear and moment diagram calculator](https://webstructural.com/shear-and-moment-diagram.html) to check your own work. ## The sign convention and the rules Positive shear pushes the left side of a cut up. Positive moment bends the beam into a smile, with the bottom in tension. Loads act down. Walk along the beam from the left end: 1. **Reactions first.** Take moments about one support to find the other, then add up the vertical forces (ΣM = 0, ΣF = 0). 2. **Shear.** An upward reaction makes the shear jump up and a point load makes it jump down, each by its own size. Under a distributed load, the shear falls at a rate equal to the load: dV/dx = −w. 3. **Moment.** The slope of the moment diagram equals the shear: dM/dx = V. The change in moment between two points is the area of the shear diagram between them. 4. **Largest moment.** It is where the shear crosses zero, or at a support or point load. 5. **Check the ends.** The shear and moment both come back to zero just past the last support or free end. If they don't, a reaction or an area is wrong. ## Example 1: two point loads A 24 ft simply supported beam carries 12 kips at 6 ft and 8 kips at 16 ft from the left support, A. **Reactions.** Moments about the right support, B: R_A = (12 × 18 + 8 × 8) / 24 = 280 / 24 = 11.67 kips R_B = 20 − 11.67 = 8.33 kips **Shear.** Between point loads nothing else acts, so the shear is flat. It is +11.67 kips from A to 6 ft, then 11.67 − 12 = −0.33 kips to 16 ft, then −0.33 − 8 = −8.33 kips to B, where R_B brings it back to zero. **Moment.** Flat shear means straight lines in the moment diagram. Add up the shear areas: M at 6 ft = 11.67 × 6 = **70.0 kip-ft**, and M at 16 ft = 70.0 − 0.33 × 10 = 66.7 kip-ft. From there, 66.7 − 8.33 × 8 = 0 at B. The shear changes sign at the 12 kip load, so the largest moment is there. The two moments are close because the shear between the loads is nearly zero. [Drawing: Shear and moment diagrams for a 24 ft simply supported beam with a 12 kip load at 6 ft and an 8 kip load at 16 ft] ## Example 2: a cantilever with a uniform load and an end load A 10 ft cantilever, fixed at the left, carries 0.5 kip/ft along its length and 3 kips at the free end. **Reactions.** The wall takes all the load: R = 0.5 × 10 + 3 = 8 kips up, and a moment of 0.5 × 10 × 5 + 3 × 10 = 55 kip-ft. **Shear.** +8 kips at the wall, falling 0.5 kip per foot to 8 − 5 = +3 kips at the free end, where the 3 kip load takes it to zero. **Moment.** The moment is **−55 kip-ft** at the wall. A cantilever bends into a frown, with the top in tension, so its moment is negative. The moment rises to zero at the free end. Check it with the shear area: (8 + 3) / 2 × 10 = 55 kip-ft. Halfway out, M = −(0.5 × 5² / 2 + 3 × 5) = −21.25 kip-ft. For a cantilever it is often easier to walk in from the free end, since no reactions are needed: M(x) = −(w x² / 2 + P x), with x measured from the free end. [Drawing: Shear and moment diagrams for a 10 ft cantilever with a 0.5 kip/ft uniform load and a 3 kip load at the free end] ## Example 3: a beam with an overhang A beam rests on supports 20 ft apart and runs 6 ft past the right one, for 26 ft in all. It carries 2 kips/ft along the whole length. **Reactions.** The total load is 2 × 26 = 52 kips, acting at 13 ft. Moments about A: R_B = 52 × 13 / 20 = 33.8 kips R_A = 52 − 33.8 = 18.2 kips **Shear.** +18.2 kips at A, falling 2 kips per foot. It crosses zero at x = 18.2 / 2 = 9.1 ft and reaches 18.2 − 40 = −21.8 kips just left of B. R_B lifts it to −21.8 + 33.8 = +12 kips, which is the load on the overhang, 2 × 6, and it falls to zero at the free end. **Moment.** The largest positive moment is where the shear crosses zero, at 9.1 ft: M = 18.2 × 9.1 / 2 = **82.8 kip-ft**, the area of the shear triangle. Over B the overhang gives a negative moment: M = −2 × 6² / 2 = **−36 kip-ft**. Between the two, the moment passes through zero at 18.2 ft, where 18.2x − x² = 0. That is the inflection point: left of it the bottom of the beam is in tension, right of it the top is. A steel beam with its top flange braced only by the floor may need the bottom flange braced near B too, and a concrete beam needs top bars over B. [Drawing: Shear and moment diagrams for a beam on supports 20 ft apart with a 6 ft overhang and a 2 kip/ft uniform load] ## Example 4: a triangular load An 18 ft simply supported beam carries a load that rises from zero at the left support to 3 kips/ft at the right, as soil or water pressure does. **Reactions.** The total load is the area of the triangle, 3 × 18 / 2 = 27 kips, acting at its centroid, two thirds of the way along, 12 ft from A: R_A = 27 × 6 / 18 = 9 kips R_B = 27 − 9 = 18 kips **Shear.** At x, the load is w = 3x / 18 = x / 6 kips/ft, and the load to the left of x is the triangle x × (x / 6) / 2 = x² / 12. So V = 9 − x² / 12, a parabola from +9 kips to −18 kips. **Moment.** V = 0 at x² = 108, so x = 10.39 ft, which is L / √3. There: M = 9 × 10.39 − 10.39³ / 36 = 93.5 − 31.2 = 62.4 kip-ft This matches the textbook formula M_max = wL² / (9√3) = 3 × 18² / 15.59 = 62.4 kip-ft. The moment diagram is a cubic, and its peak sits past midspan, toward the heavy end. See the [beam deflection formulas](https://webstructural.com/beam-deflection-formulas.html) for this case's deflection. [Drawing: Shear and moment diagrams for an 18 ft simply supported beam with a load rising from zero to 3 kip/ft] ## Example 5: an applied moment A 16 ft simply supported beam has a 40 kip-ft clockwise moment applied at 10 ft, as from a bracket or a column bolted to its side. No other load acts. **Reactions.** The supports have to make a couple that cancels the 40 kip-ft: R × 16 = 40, so R = 2.5 kips, down at A and up at B. **Shear.** A moment adds no vertical force, so the shear is −2.5 kips the whole way. The shear diagram doesn't show the moment at all. **Moment.** From A, M = −2.5 x, reaching −25 kip-ft just left of 10 ft. A clockwise applied moment makes the moment diagram jump up by its size: −25 + 40 = +15 kip-ft. From there the moment falls at 2.5 kip-ft per foot to 15 − 2.5 × 6 = 0 at B. The largest moment, 25 kip-ft, is just left of the applied moment, and the shear never crosses zero. The rule "largest moment where the shear is zero" misses this case, so always check the moment on both sides of an applied moment. [Drawing: Shear and moment diagrams for a 16 ft simply supported beam with a 40 kip-ft clockwise moment applied at 10 ft] ## Common mistakes - **Placing a distributed load at the wrong point.** A uniform load acts at its middle, a triangle at two thirds of the way toward its tall end. - **Missing the second sign change.** An overhang has a second place where the shear crosses zero, at the support, and a negative moment there that can govern. - **Mixing feet and inches.** Keep kips and feet through the diagrams, then turn kip-ft into kip-in (× 12) for stresses. - **Ignoring a mismatch at the end.** If the shear doesn't come back to zero, recheck the reactions before going on. ## Check it with the calculators - The [shear and moment diagram calculator](https://webstructural.com/shear-and-moment-diagram.html) draws the reactions, shear, moment and deflection for any of these beams, and for continuous beams over several supports. - The [beam calculator](https://webstructural.com/beam-calculator.html) gives the equations for 17 single-span load cases. - The [beam designer](https://webstructural.com/beam-designer.html) takes the moment and shear and picks a steel or wood beam that is strong and stiff enough. The [steel beam design tutorial](https://webstructural.com/steel-beam-how-to.html) shows those checks by hand. ## Frequently asked questions ### How do you draw a shear and moment diagram? Find the reactions, then walk along the beam from the left. Point loads make the shear jump, and distributed loads make it slope. The moment at any point is the area of the shear diagram up to that point. ### Where is the maximum bending moment? Where the shear crosses zero, or at a support, a point load or an applied moment. Check each of them. ### Why is the moment negative over a support? An overhang or the next span bends the beam into a frown over the support, putting the top in tension. That is negative moment. ### What is the relationship between shear and moment? The shear is the slope of the moment diagram, dM/dx = V, and the load is the slope of the shear diagram, dV/dx = −w. More worked examples: [browse all structural design tutorials](https://webstructural.com/structural-learning-hub.html).