Question

It's not possible to directly measure the yield strength for steel. 2) Find the stress and strain in each rod and total elongation of the bar shown in Figure The yield and ultimate strengths of a material. Explain P$_2$=40,000 N P$_3$=90,000 N P$_4$=150,000 N P$_1$=100,000 N $\diamond$35 $\diamond$25 $\diamond$50 L$_s$=600 L$_b$=400 L$_c$=350 Steel, E$_s$=200 Gpa Bronze, E$_b$=84 Gpa Copper, E$_c$=120 Gpa

          It's not possible to directly measure the yield strength for steel.
2) Find the stress and strain in each rod and total elongation of the bar shown in Figure
The yield and ultimate strengths of a material. Explain
P$_2$=40,000 N
P$_3$=90,000 N
P$_4$=150,000 N
P$_1$=100,000 N
$\diamond$35
$\diamond$25
$\diamond$50
L$_s$=600
L$_b$=400
L$_c$=350
Steel, E$_s$=200 Gpa
Bronze, E$_b$=84 Gpa
Copper, E$_c$=120 Gpa

It's not possible to directly measure the yield strength for steel.
2) Find the stress and strain in each rod and total elongation of the bar shown in Figure
The yield and ultimate strengths of a material. Explain
P2=40,000 N
P3=90,000 N
P4=150,000 N
P1=100,000 N
♢35
♢25
♢50
Ls=600
Lb=400
Lc=350
Steel, Es=200 Gpa
Bronze, Eb=84 Gpa
Copper, Ec=120 Gpa

Added by Jose Luis H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Thanks for your feedback!

It is not possible to directly measure the yield strength of steel. The ultimate strength of a material can be determined instead. P = 100,000 N P = 40,000 N P = 90,000 N P = 150,000 N P = $35 P = $25 PA = 50 Ly = 600 Ln = 400 Lc = 350 Steel, E = 200 GPa Bronze, Ey = 84 GPa Copper, Ec = 120 GPa