Civil Engineering

Question

The diameter of the brass and steel segments of the axially loaded bar shown in fig. are 30 mm and 12 mm respectively. The diameter of the hollow section of the brass segment is 20 mm. Determine (i) The maximum normal stress in the steel and brass (ii) The displacement of the free end. Take Es=210 GN/m2, Eb= 105 GN/m2.[Ans: σs=88.42MN/m2, σb(BC)=707MN/m2,σb(BC)=13.7MN/m2, Ϩl=0.092mm]

link to figure: https://ibb.co/b58Gdg9

Please help me guys, tomorrow I've an exam and this question has a high chance to appear...!

Answer 1

Armit K. I just saw your problem and sorry I missed your exam, but if any consolation to you, here is the solution you are looking for:

You need the stress in each of the cylinders in the steel section and in the brass section which consists of two sections one solid and one hollow

1- Steel section: This is solid section of diameter 12 mm between A and B

Let As_(AB) be area of steel section with length of 0.15 m between A and B

As_(AB) = π (12 mm)² /4 = 36π mm²

Taking a free body diagram with a section through the steel cylinder between A and B, The axial force is

F_{s (AB)} = 10 KN

The stress in the steel section

σs = F_s(AB) / /As_(AB) = 10 KN/ 36π mm² = 10 x 10³ N / 36π ( 10^-3)² m² =88.42 x 10⁶ N/m²

= 88.42 MN / m²

2- Brass section: This is solid section of diameter 30 mm between B and C

Let A_b(BC) be area of brass section with length of 0.2 m between B and C

A_b(BC) = π (30 mm)² /4 = 225π mm²

Taking a free body diagram with a section through the brass Cylinder between B and C, The sum of axial force is -10 KN to the left and 5 KN to the right for a net force of 5 KN

F_{b (BC)} = 5 KN

The stress in the brass section between B and C is

σ_b(BC) = F_b(BC) / A_b(BC) = 5 KN / 225π mm² = 5 x 10³ N / 225π ( 10^-3)² m² =7.07 x 10⁶ N/m²

= 7.07 MN / m²

Note: I believe that the answer above is a typo error it should be 7.07 rather than 707

2- Brass section: This is a hollow section of outside diameter of 30 mm and inside diameter of 20 mm between C and D

Let A_b(CD) be area of brass section with length of 0.125 m between C and D

A_b(CD) = π [(30 mm)² - (20 mm)²] /4 = 125π mm²

Taking a free body diagram with a section through the brass cylinder between C and D, The equilibrium of internal axial force indicates 5 KN to the left

F_{b (CD)} = 5 KN

The stress in the brass section between C and D is

σ_b(CD) = F_b(CD) / A_b(CD) = 5 KN / 125π mm² = 5 x 10³ N / 125π ( 10^-3)² m² = 12.7 x 10⁶ N/m²

= 12.7 MN / m²

Note: Again I believe that the answer is 12.7 not 13.7 as shown above

4) I guess the uncertainty in answers can be tested when we find the displacement, if we calculate the displacement and it comes accurate as shown above as 0.092 mm, that validates the solution to the above calculated stresses which are part of the displacement equation, so lets see

The displacement at the free end can be approximated by δ = ∑P_i L_i / A_i E_i = ∑ σ_i ( L_i / E_i) where σ_i is the individual stresses in the section as calculated above

δ_s(AB) = 88.42 x 10⁶ N/m² ( 0.15 m) / 210 x 10⁹ N / m² = 0.06315 x 10^-3 m

δb_(BC) = 7.07 x 10⁶ N/m² ( 0.2 m) / 105 x 10⁹ N / m² = 0.01347 x 10^-3 m

δb_(CD) = 12.7 x 10⁶ N/m² ( 0.125 m) / 105 x 10⁹ N / m² = 0.01511 x 10^-3 m

Total Displacement at free end = δ_s(AB) + δb_(BC) + δb_(CD) = ( 0.06315+0.01347+0.01511) x 10^-3 m =

0.09173 x 10^-3 m x ( 10 ³ mm / m ) = 0.09173 mm = 0.092 mm same as answer given, and the stresses are therefore are accurate as calculated.