Question

A bar of length 3 m has a diameter of 50 mm over half its length and a

diameter of 25 mm over the other half. If E= 2.06 x 105 N/mm2

and the bar

is subjected to a pull of 50 kN. Find the stress in each section and total

extension of the bar.​

Answer 1

Answer:

BHAI PLAEE MARK AS BRAINLIEST ANSWER

Answer 2

Answer:

stress along first half=2.54 × $10^{7}$N/$m^{2}$

stress along second half=10.16 × $10^{7}$N/$m^{2}$

total extension of the bar=0.1875m

Explanation:

the deformation of a bar with known geometry and subjected to a load can be determined by the equation

δ=FL/AE

AE=axial rigidity

L=length of the material

Area=π x radius x radius

case I : along first half

radius=diameter/2=50 × $10^{-3}$/2=25 × $10^{-3}$

Stress=F/A=50 × $10^{3}$/π ×(25×25) x $10^{-6}$         

          =2.54 x $10^{7}$N/$m^{2}$

δ1=stress × L/E= 2.54 × $10^{7}$ × 3/2.06 × $10^{9}$

δ1=3.69 × $10^{-2}$=0.0396m

case II: along the other half

radius=diameter/2=25 × $10^{-3}$/2=12.5 × $10^{-3}$

Stress=F/A=50 × $10^{3}$/π ×(12.5 × 12.5)$10^{-6}$

         =10.16 × $10^{7}$N/$m^{2}$

δ2=stress x L/E=10.16 × $10^{7}$× 3/2.06 × $10^{9}$

δ2=14.79 × $10^{-2}$=0.1479m

total extension of the bar=δ1+δ2=0.1479+0.0396=0.1875m