Question

Axial (Normal) Stress, Strain and Deformation The system is loaded and supported as shown. The system is made of Brass, E, 98.9 GPa, is a solid shaft and has a diameter of d 27 mm. 9 kN 10 kN A D B 6 kN E 15 kN 9 kN 10 kN + En: 2010) 629) 145 mm 85 mm 50 mm 60 mm See hon in cd P tol15 P-a3 -33-0.056 T-P/A Calculate the axial normal stress in section CD, ocp- (4 pts) 1) 57.6 M 61.8 MPa (T) b. Pa (T) a. 64.0 MPa (C) 59.7 MPa (C) C. d. Calculate the displacement of point D relative to point C. (4 pts) 2) 0.0867 mm b. 0.0913 mm a. d. 0.0845 mm 0.0811 mm C. (4 pts) Calculate the axial normal strain in section AB, ean 3) b. 119 (10)/m 129 (10mm/ a. mm. 124 (10m/m d. 114 (10)/ C

Answer 1

Ans) Given ,

Modulus of elasticity of Brass (E) = 98.9 GPa or 98900 N/mm²

Diameter (d) = 27 mm

1) Net force in bar CD ,

P_CD = 15 + 2(9) = 33 kN

We know, $\dpi{100} \sigma$ _CD = P_CD / A

= 33000 / [($\dpi{100} \pi$/4) x 0.027 x 0.027]   

= 57.66 x 10⁶ N/m²

= 57.66 MPa

2) We know,

  $\dpi{100} \delta$_D = P_CD L_CD / A E

= 33000 x 145 / [($\dpi{100} \pi$/4) x 27 x 27 x 98900]

= 0.0845 mm

3) Axial normal strain in AB ,

($\dpi{100} \delta$/L)_AB = P_AB / A E

P_{AB =} 15 + 2(9) - 2(10) - 6

= 7 kN

Putting values,

($\dpi{100} \delta$/L)_AB = 7000 / [($\dpi{100} \pi$/4) x 27 x 27 x 98900]

= 1.24 x 10^-4 mm/mm

or 124 x 10^-6 mm/mm