Help me with my machine design homework! During a tension test, a mild steel spe

Question

Help me with my machine design homework!

During a tension test, a mild steel specimen of diameter 12 mm and gage length 60 mm elongates to 75 mm. The rod can sustain a maximum load of 50 kN yields at 25 kN and breaks at 30 kN. Find its yield strength, ultimate strength, strength at the point of failure, actual strength at the point of failure when the diameter is reduced to 8 mm, percentage elongation, and percentage reduction in area. Show the calculated strength in the stress-strain curve. (20)

Explanation / Answer

The cross-sectional area, for r =12 mm = pi* r *r = 4.52 X 10^-4 m^2

The cross-sectional area, for r =8 mm = 2 X 10 ^-4 m^2

So, yield strength = yielding force / Original Area = 25000/ ( 4.52 X 10^-4 ) = 55.3 MPa

Ultimate strength = Maximum load / Original area = 50000/ ( 4.52 X 10^-4 ) = 110.6 MPa

Strength at failure = Breaking Load / Area at the time of failure = 30000/( 2 X 10^-4) = 150 MPa

Percentage reduction in area = ( 4.52 - 2) *100 /4.52 = 55.7 %

percentage elongation in length = (75-60) *100 /60 = 25 %

The above values can be entered in the stress strain curve of a mild steel for the corresponding points points on the graph.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.