Question

Section 2 (50 marks) QUESTION 4 (20 marks) As a project engineer you have been requested to select an ap steel arch bridge with 100m span. The metal selected from Table 2 needs to h criteria met: propriate metal to be used as bridge girders of a ave the following two performance i) Sufficient ductility i) High strength Table 2: Laboratory test results of Metals Final gauge length (mm) 51.08 59.75 59.28 68.53 Peak l Original diameter (mm) 10.161 10.220 10.126 10.013 Final diameter Fracture load (kN) 23.95 54.59 25.61 26.56 Metal Metal A Metal B Metal C Metal D 10.061 7.253 7.605 6.670 (kN) 23.9 79.70 31 38.04 Note: Original gauge length of the specimens is 50mm

Answer 1

Metal A :-

ultimate strength = (Peak load) / (final cross sectional area) = $\frac{23.95\times 10^{3}}{\frac{\pi}{4}\times 10.061^{2}} = 301.25 MPa$

and strain of the gauge = $\Delta L/L=(51.08-50)/50 = 0.0216 =2.16$ %

Metal B :-

ultimate strength = (Peak load) / (final cross sectional area) = $\frac{79.70\times 10^{3}}{\frac{\pi}{4}\times 7.253^{2}} = 1929 MPa$

and strain of the gauge = $\Delta L/L=(59.75-50)/50 = 0.195 =19.5$ %

Metal C :-

ultimate strength = (Peak load) / (final cross sectional area) = $\frac{31.90\times 10^{3}}{\frac{\pi}{4}\times 7.605^{2}} = 702.27 MPa$

and strain of the gauge = $\Delta L/L=(59.28-50)/50 = 0.1856 =18.56$ %

Metal D :-

ultimate strength = (Peak load) / (final cross sectional area) = $\frac{38.04\times 10^{3}}{\frac{\pi}{4}\times 6.67^{2}} = 1088.7 MPa$

and strain of the gauge = $\Delta L/L=(68.53-50)/50 = 0.3706 =37.06$ %

Metal	Ultimate Strain(%)	Ultimate Strength(MPa)
Metal A	2.16	301.25
Metal B	19.5	1929
Metal C	18.56	702.27
Metal D	37.06	1088.7

for the selection of metal, we have two criteria

(1) sufficient Ductility (Metal A<Metal B<Metal C<Metal D)

(2) High Strength (Metal A <Metal C<Metal D<Metal B)

so, we should choose Metal B ,as this mettle has sufficient ductility with high strength.