1.10 Introduction to 3D force systems
After this lesson you should be able to:
- Write a force in three dimensions using unit vectors, components and direction cosines.
- Compute the moment of a force about a point (r × F) and about an axis.
- State the six equilibrium equations in 3D and the reactions of common 3D supports.
Theory
Many civil engineering systems are three-dimensional: transmission towers, guyed masts, space frames, tripods for lifting, and the forces on a bridge pier from braking, wind and dead load at once. The ideas of the previous lessons carry over directly; we simply add a third axis.
A force in space is written F = Fx i + Fy j + Fz k. Its magnitude is √(Fx² + Fy² + Fz²), and the angles θx, θy, θz it makes with the axes have direction cosines l = Fx / F, m = Fy / F, n = Fz / F, with l² + m² + n² = 1. When a force acts along a line from point A to point B, the unit vector along AB is the coordinate difference divided by the length, and F = F u, exactly as in two dimensions.
In space a moment is a vector. The moment of F about point O is M_O = r × F, where r runs from O to any point on the line of action. Its components are the moments about the three axes. The moment about any axis through O with unit vector λ is the projection λ · (r × F), which measures how much the force tends to turn the body about that axis.
A rigid body in space has six independent ways to move: three translations and three rotations. Equilibrium therefore needs six equations: ΣFx = ΣFy = ΣFz = 0 and ΣMx = ΣMy = ΣMz = 0. Supports again provide one reaction per prevented movement: a ball support on a smooth surface gives one force; a ball-and-socket gives three forces; a fixed support gives three forces and three moments. For concurrent forces in space, as at a joint of a space truss, only the three force equations are needed.
Derivation
- Let F act along AB with A = (x₁, y₁, z₁) and B = (x₂, y₂, z₂). Then AB = (Δx, Δy, Δz) and d = √(Δx² + Δy² + Δz²).
- u = (Δx, Δy, Δz) / d, so F = F u and the direction cosines are Δx / d, Δy / d, Δz / d.
- Since u is a unit vector, l² + m² + n² = 1.
- With r = (x, y, z) and F = (Fx, Fy, Fz), expanding the determinant gives r × F = (y Fz − z Fy) i + (z Fx − x Fz) j + (x Fy − y Fx) k.
- The k-component, x Fy − y Fx, is the 2D moment formula of lesson 1.4: plane statics is a special case of 3D statics.
F = F u, u = (Δx, Δy, Δz) / d; F = √(Fx² + Fy² + Fz²)Direction cosines l = Fx / F, m = Fy / F, n = Fz / F; l² + m² + n² = 1M_O = r × F = (y Fz − z Fy, z Fx − x Fz, x Fy − y Fx)Moment about an axis with unit vector λ: M_λ = λ · (r × F)3D equilibrium: ΣF = 0 (3 equations), ΣM_O = 0 (3 equations)
- Rigid body; supports idealised.
- Right-handed axes: x × y = z.
- Moments obey the right-hand rule: thumb along the moment vector, fingers show the turning sense.
Codes and standards
Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.
- No design code governs this lesson. Three-dimensional analysis of towers and space structures follows the mechanics shown here; codes such as IS 802 (Part 1/Sec 1):2015 for transmission line towers and EN 1993-3-1 for towers and masts then govern loads and member design. Listed for comparison, not ranked.
Worked examples
Problem. A cable runs from A(0, 0, 0) to B(2 m, 3 m, 6 m) and carries a tension of 900 N from A towards B. Write the force in components and find its direction angles.
- AB = (2, 3, 6) m; d = √(4 + 9 + 36) = 7 m.
- u = (2/7, 3/7, 6/7).
- F = 900 u = (257.14, 385.71, 771.43) N.
- Direction angles: θx = cos⁻¹(2/7) = 73.40°, θy = cos⁻¹(3/7) = 64.62°, θz = cos⁻¹(6/7) = 31.00°.
- Check: (2/7)² + (3/7)² + (6/7)² = 49/49 = 1.
Answer. F = (257.1, 385.7, 771.4) N; angles 73.40°, 64.62°, 31.00°
Problem. A bracket on a pier face receives a force F = (2, −3, 4) kN at the point r = (1, 2, 0) m from the pier centre O. Find the moment vector about O, its magnitude, and the moment about the x-axis.
- M_O = r × F = (y Fz − z Fy, z Fx − x Fz, x Fy − y Fx).
- Mx = 2 × 4 − 0 × (−3) = 8 kN·m; My = 0 × 2 − 1 × 4 = −4 kN·m; Mz = 1 × (−3) − 2 × 2 = −7 kN·m.
- |M_O| = √(8² + 4² + 7²) = √129 = 11.36 kN·m.
- The moment about the x-axis is the x-component: 8 kN·m.
Answer. M_O = (8, −4, −7) kN·m; magnitude 11.36 kN·m; about x-axis 8 kN·m
Step-by-step procedure
- Set up right-handed axes and write coordinates of all key points.
- Express each force as magnitude times a unit vector along its line.
- Write position vectors from the moment centre to a point on each force.
- Compute cross products term by term (or by determinant) and sum.
- Apply the six equilibrium equations, or three for a concurrent system.
Site notes
- Guy wires on masts and temporary towers are 3D force systems: each guy's tension has a vertical component that adds to the mast compression. Set-out errors in anchor positions change these components.
- Braking and wind forces on bridge piers act in different directions; the pier base sees a 3D set of forces and moments, which is why pier foundations are checked about both axes.
- When lifting with a three-legged spreader or tripod, unequal leg lengths make the leg forces unequal; work them out with unit vectors rather than assuming equal shares.
- Using a left-handed axis system, which flips moment signs.
- Getting the order wrong in r × F (F × r gives the negative).
- Forgetting that r must run from the moment centre to a point on the force's line.
- Treating a 3D problem as 2D by ignoring one component.
- Expecting more than three unknowns to be found from a concurrent 3D force system.
Direction cosines always satisfy l² + m² + n² = 1, and each lies between −1 and 1. A force along a cable whose vertical rise is the largest dimension will have its largest component vertical.
Summary
- F = F u with u from coordinates; l² + m² + n² = 1.
- M_O = r × F; components are the moments about the axes.
- Moment about an axis = λ · (r × F).
- Six equilibrium equations in 3D; three for concurrent forces.
Practice questions
Tap a question to see the answer.
1What is the sum of the squares of the direction cosines of any force?
1.
2Find the magnitude of F = (3, 4, 12) kN.
√(9 + 16 + 144) = 13 kN.
3Compute r × F for r = (0, 0, 2) m and F = (5, 0, 0) kN.
(0 × 0 − 2 × 0, 2 × 5 − 0 × 0, 0 × 0 − 0 × 5) = (0, 10, 0) kN·m.
4How many independent equilibrium equations does a rigid body have in 3D?
Six: three force and three moment equations.
5Find the unit vector from (1, 1, 1) to (3, 3, 2).
(2, 2, 1) / 3 = (0.667, 0.667, 0.333).
6A force F = (0, 0, −10) kN acts at (2, 3, 0) m. Find its moment about the x-axis.
Mx = y Fz − z Fy = 3 × (−10) − 0 = −30 kN·m.
References
Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.
- F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
- J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
- S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.