81/3848–12FE Reference Handbook 10.6 · Mechanics of Materials · p. 135: w(x) = −dV/dx and V = dM/dx. Not printed in the Statics chapter.
1. Moment & Equilibrium2. Centroid of a Composite Area3. Coulomb Friction4. Force Resultant5. Rectangle Centroidal Moment of Inertia6. Parallel-Axis Theorem7. Distributed Load Resultant8. Method of Joints9. Two-Force Member10. Ideal Pulley11. Moment of a Force12. Simply Supported Beam Reactions13. Load–Shear–Moment Relations14. Method of Sections (Truss)15. Sagging Cable (Small Sag)16. Overhanging-Beam Reactions17. Radius of Gyration18. Triangle Area Moment of Inertia19. Triangular Distributed Load Resultant20. Three-Force Member21. Couple Moment22. Frames: Multi-Force Members23. Zero-Force Truss Members

Handbook formula

Load–Shear–Moment Relations

Shear and moment signs are printed in Mechanics of Materials p. 135, not in Statics: positive moment bends the beam concave upward, and positive shear means the right portion tends to move down relative to the left. With that convention w(x) = −dV/dx and V = dM/dx. A concentrated force jumps V; a concentrated couple jumps M.

(kN/m)
Distributed load
(kN)
Shear
(kN·m)
Moment

Step-by-step solved example

Simple beam L=4 m, w=2 kN/m uniform. V(x) from left and Mmax?

w = 2 kN/mV = 4V = −4Mmax = 4 kN·mV(x) = 4 − 2x kNMmax = wL²/8 = 4 kN·mReplay
Load → shear → moment by integration. dV/dx = −w, dM/dx = V.

  1. 1. Reactions

    RA=RB=4 kN.

  2. 2. V and M

    V(x)=4−2x kN. V=0 at x=2 m. Mmax=4×2−2×2×1=4 kN·m = wL²/8.

Answer: V=4−2x kN; Mmax=4 kN·m

10 practice questions

0/10 correct

1.dM/dx equals

2.dV/dx equals

3.A downward point load causes V to

4.Mmax on a simple uniform beam is

5.Where V=0 (no point moment), M is

6.Cantilever with end load P: V is

7.A couple M0 on the beam makes M

8.Uniform w=3 kN/m, L=6 m simple. RA=

9.If w=0 between two points, V is

10.M diagram slope is