When a beam carries a load, internal forces develop to keep it from breaking apart. Two forces matter most: shear (the internal force acting perpendicular to the beam's axis, trying to slide one section past another) and bending moment (the internal torque trying to bend the beam). Plotting these along the beam's length gives you shear and moment diagrams — the single most important tool for sizing beams in structural and mechanical design.

The simply supported beam is the classic starting case: a beam resting on two supports with a load between them. Consider a 4-meter beam supported at both ends with a single 10 kN point load at its center (2 m from each support).

• Step 1 — Reactions: By symmetry, each support carries half the load. R_A = R_B = 5 kN (upward).
• Step 2 — Shear diagram: Starting from the left, shear jumps to +5 kN at the support, stays constant (no other loads yet), then drops by 10 kN at the center load to −5 kN, and stays at −5 kN until the right support brings it back to zero. The diagram looks like a step function.
• Step 3 — Moment diagram: Moment is the running integral of the shear diagram. It starts at zero, increases linearly to a peak of 5 kN × 2 m = 10 kN·m at midspan, then decreases linearly back to zero. The shape is a triangle.

That peak moment of 10 kN·m is what you'd use to select a beam. You compare it against the beam's section modulus (S) and the material's allowable stress (σ): M = σ × S. For A36 structural steel with an allowable bending stress of roughly 150 MPa, you need S ≥ 10,000 N·m / 150 × 10⁶ Pa = 66.7 × 10⁻⁶ m³, or about 66.7 cm³. A standard W150×13 wide-flange beam (S ≈ 72 cm³) would work.

Critical rules to internalize:

• Shear is the derivative of the applied distributed load; moment is the integral of shear.
• Maximum moment occurs where shear crosses zero.
• Point loads cause jumps in the shear diagram; distributed loads cause slopes.
• Moment diagrams are one degree higher than shear: if shear is constant, moment is linear; if shear is linear, moment is parabolic.

Real-world example: A floor joist in residential construction is essentially a simply supported beam carrying a distributed load (furniture, people, the subfloor). Building codes specify a live load of about 1.9 kPa (40 psf). Engineers draw the shear and moment diagrams for the joist span, find the peak moment, and select the lumber size — this is exactly why a 2×10 is required for a 14-foot span but a 2×8 suffices for 10 feet.