Stress: force inside the material
Stress tells how hard the material is being pushed or pulled. It is force spread over an area: σ = F / A. Unit: N/mm² = MPa.
Pull or push along the bar gives normal stress. Sliding forces give shear stress τ = V / A. Bending makes one side of a beam squeezed and the other side stretched: σ = M / Z, where Z is the section modulus (more on this in the lesson on members).
The same force on a thinner bar gives more stress. That is why thin bars break first.
What does statically determinate mean?
A body at rest has three equilibrium equations in a plane: ΣFx = 0, ΣFy = 0, ΣM = 0. If a structure has exactly three unknown support reactions (or the unknowns equal the equations), they can all be found by these equations alone. We call it statically determinate.
Supports: a roller gives 1 reaction, a pin gives 2, a fixed support gives 3.
Statically determinate beams
Simply supported beam of span L with a point load P at distance a·L from support A:
- R_B = P·a, R_A = P(1 − a) (take moments about A, then about B).
- Bending moment under the load: M = P·a·(1 − a)·L. It is zero at both supports.
- Even load w on the whole span: R = wL/2 each, M_max = wL²/8 at mid-span.
- Cantilever (fixed at one end) with tip load P: M at the fixed end = P·L, shear = P everywhere.
Shear force jumps at every point load; bending moment changes slope there.
Statically determinate frames
A frame is made of beams and columns joined at corners. A frame with one pin and one roller support has three reactions, so it is determinate. Steps: (1) draw the free-body, (2) get the three reactions from ΣFx, ΣFy, ΣM, (3) cut the frame at any section and find the axial force N, shear V and moment M there.
A side push on a portal frame makes the two columns bend; the moment at the top corner passes from column to beam, so the corner must be strong.
Statically determinate trusses
A truss has straight bars joined at pins, loads act only at the joints, so each bar carries only axial force: tension (pull, +) or compression (push, −).
It is determinate when m + r = 2j (m bars, r reactions, j joints). Two ways to solve: method of joints (balance forces at each pin) and method of sections (cut through three bars and balance one part).
Triangle truss, load P at the top: R = P/2; sloping bar force = (P/2)/sin θ (compression); bottom bar = (P/2)/tan θ (tension).
Key formulas and definitions
- ΣFx = 0, ΣFy = 0, ΣM = 0
- σ = F / A (N/mm² = MPa)
- Simply supported, point load: R_A = P·b/L, R_B = P·a/L
- M_max (point load) = P·a·b / L; M_max (even load) = wL² / 8
- Cantilever: M = P·L at the fixed end
- Truss determinate if m + r = 2j
- Triangle truss: side bar = (P/2)/sin θ, bottom bar = (P/2)/tan θ
Worked examples
1. A 6 m simply supported beam carries a 12 kN load 2 m from support A. Find R_A, R_B and the moment under the load.
b = 4 m. R_A = P·b/L = 12 × 4 / 6 = 8 kN. R_B = 12 × 2 / 6 = 4 kN. Check: 8 + 4 = 12. Moment under load = R_A × 2 = 16 kN·m.
2. An 8 m beam carries an even load of 5 kN/m. Find the reactions and the maximum bending moment.
Total load = 5 × 8 = 40 kN. R = 20 kN each. M_max = wL²/8 = 5 × 64 / 8 = 40 kN·m at mid-span.
3. A cantilever 3 m long has a 4 kN load at its free tip. Find the moment and shear at the fixed end.
Shear = 4 kN. Moment = P × L = 4 × 3 = 12 kN·m.
4. A bar of 20 mm diameter pulls with 31.4 kN. Find the stress.
A = π × 10² = 314 mm². σ = 31 400 / 314 = 100 MPa.
5. A triangle truss has half-span 3 m and height 4 m. A 24 kN load acts at the top joint. Find the force in each bar.
Slope bar length = 5 m, sin θ = 4/5, tan θ = 4/3. Each support = 12 kN. Sloping bars: 12 / 0.8 = 15 kN compression. Bottom bar: 12 / (4/3) = 9 kN tension.
6. A truss has 7 bars, 5 joints and 3 support reactions. Is it statically determinate?
m + r = 7 + 3 = 10 and 2j = 10. They are equal, so it is determinate.
Common mistakes
- Taking moments about a point but forgetting one of the forces (a reaction through that point has zero moment, which is why we choose it).
- Mixing kN and N or m and mm in σ = F/A. Write units on every line.
- Calling a bar in tension negative (or the other way) without keeping one sign rule for the whole problem.
- Thinking the beam needs a different rule when the load moves. The rules stay the same; only the numbers change.