A solid rod of 12 mm diameter was tested for tensile strength with the gauge length of 50 mm. final length = 80 mm; final diameter = 4 mm; yield load = 1130 n. what is the nearest yield stress and % reduction in area respectively?
Answers
Answered by
2
Yield stress = ultimate force / original cross - sectional area
Convert the measurements of the diameters into metres.
1000mm = 1m
Original diameter = 12/1000= 0.012m
Radius = 0.006m
Final diameter = 4/1000=0.004
Radius = 0.002
Original cross sectional area: 3.142 × 0.006² = 1.13112 × 10⁻⁴ m²
Final cross sectional area : 3.142 × 0.002² = 1.2568 × 10⁻⁵m²
Yield stress = 1130N / 1.13112 × 10⁻⁴ = 9.551082231 × 10⁶ N/M²
The percentage reduction :
1.131120× 10⁻⁴ - 1.2568 ×10⁻⁵ = 1.00544 × 10⁻⁴
1.00544 × 10⁻⁴ / 1.13112 ×10⁻⁴ = 0.88888
% reduction : 0.8888 × 100% = 88.89%
Convert the measurements of the diameters into metres.
1000mm = 1m
Original diameter = 12/1000= 0.012m
Radius = 0.006m
Final diameter = 4/1000=0.004
Radius = 0.002
Original cross sectional area: 3.142 × 0.006² = 1.13112 × 10⁻⁴ m²
Final cross sectional area : 3.142 × 0.002² = 1.2568 × 10⁻⁵m²
Yield stress = 1130N / 1.13112 × 10⁻⁴ = 9.551082231 × 10⁶ N/M²
The percentage reduction :
1.131120× 10⁻⁴ - 1.2568 ×10⁻⁵ = 1.00544 × 10⁻⁴
1.00544 × 10⁻⁴ / 1.13112 ×10⁻⁴ = 0.88888
% reduction : 0.8888 × 100% = 88.89%
Similar questions