Tie bar that 'did not break' but stretched too far
A roof truss fabricated on site uses a 1.20 m long, 20 mm diameter mild steel tie bar with forged eyes at both ends. The design data are: working tensile load 60 kN, yield strength of the supplied bar 250 MPa, elastic modulus 200 GPa, and a drawing requirement of factor of safety 2.5 on yield. During proof loading the bar carried the full 60 kN, and its extension read on a dial gauge over the full 1.20 m was 1.80 mm. The bar showed no necking and returned to its original length when the load was removed. The fabricator's inspector signed it off as satisfactory on the ground that it did not break. The site engineer must accept or reject the truss before roof sheeting begins, and the only bar stock in the yard is 25 mm diameter in 1.2 m lengths.
1. Calculate the working stress and the actual factor of safety on yield. Is the bar acceptable as designed?
Read the worked answer
Cross-sectional area: A = πd²/4 = π × 20²/4 = 314.2 mm².
Working stress: σ = P/A = 60,000 N / 314.2 mm² = 191 N/mm² (MPa).
Permissible stress from the drawing: σ_allow = σ_y / FoS = 250 / 2.5 = 100 MPa.
Actual factor of safety: FoS_actual = 250 / 191 = 1.31.
Verdict: reject. The bar is overstressed by a factor of about 1.9 relative to the drawing requirement. It survives the proof load only because 191 MPa is below the 250 MPa yield, but it has almost no margin for load sharing error, wind, erection loads, temperature, or a slight underrun in diameter. A 1.31 factor of safety on yield is below what any structural tie should carry, and the absence of necking proves nothing - the bar is elastic, so no permanent set is expected at 191 MPa. The "it did not break" sign-off is not a valid acceptance criterion; the criterion is stress against the permissible value.
2. Back-calculate the apparent elastic modulus from the measured extension and interpret the result. What would you check before concluding that the steel is non-conforming?
Read the worked answer
From δ = PL/AE, E = PL/(Aδ):
E = (60,000 × 1200) / (314.2 × 1.80) = 72,000,000 / 565.6 ≈ 127,000 N/mm².
Expected elastic extension for steel: δ = (60,000 × 1200)/(314.2 × 200,000) = 1.15 mm. The measured 1.80 mm is about 57% high.
Interpretation: an apparent modulus of 127 GPa is far below the 200–210 GPa of any structural steel. No normal mild steel behaves this way, so the most probable explanations are measurement or setup faults, not a magical low-modulus steel:
- Gauge datum and slip - the dial gauge was probably bearing on the eye or on the loading shackle, so clevis take-up and pin seating are included in the 1.80 mm. Real extension should be measured on the shank only, between punch marks.
- Load path - the applied load may exceed 60 kN if the load cell or the proving arrangement is out of calibration, or the bar may have been proof-loaded through a lever with an unknown ratio.
- Temperature and straightening - thermal expansion from sun on the bar, or straightening of an initially bowed bar, adds apparent extension.
- Section - check the actual diameter at several points; a bar rolled under size (say 18 mm) raises stress and elongation together.
Only after eliminating these should a tensile test be cut from the bar to confirm yield and modulus.
3. The yard only has 25 mm bar. Check whether 25 mm is sufficient, and if not, state the minimum diameter and the practical consequence for the end connections.
Read the worked answer
Area of 25 mm bar: A = π × 25²/4 = 490.9 mm².
Stress at 60 kN: σ = 60,000 / 490.9 = 122 MPa.
Permissible stress is 100 MPa, so 25 mm is still not sufficient - FoS on yield would be 250/122 = 2.05, short of the required 2.5.
Required area: A_req = P/σ_allow = 60,000/100 = 600 mm².
Required diameter: d = √(4A/π) = √(4 × 600/π) = √763.9 = 27.6 mm, so specify 28 mm as the next available standard size (or redesign the member so the load is shared by two bars).
Practical consequence: going from 20 mm to 28 mm is a 40% increase in diameter, which changes the whole connection. The forged eyes, clevis pins and gusset bolt holes must be re-sized; a 28 mm eye will not pass the existing pin, and simply drilling out the truss gusset for a larger pin may reduce the net section of the gusset below its own allowable stress. Either way, the change should go back to the designer as a revision, not be decided in the yard. If time is critical, a temporary propping scheme can secure the roof, but the tie bar must not be sheeted over until it is replaced.