Question

3 نقطة (نقاط) السؤال 1 A bar of 20 mm diameter is made of a material having a modulus of elasticity of E= 89 GPa and a modulus of rigidity of G = 61 GPa. If it is loaded axially with a compressive force of 65 kN, determine the absolute .change in its diameter التالي

Answer 1

solution :- xy d~20mm De 65 KN 65KN Given that diame tee of bar d=20mm = 0.02m axial compressive force p= 65KN. modulus of elasticity E= 89 Gpa. modulus of Cross-sectional area (A) = F/4 d² rigidity G = 61 GPa now = 7/4 (0.02m) ² A=0.0003141 m2 -65KN 0.000314 m2 aug. normal stress o=P/A (: N/m² pa. 5= 206.94 MPa. Compressive stress 4. considering the material behaver elastically. - 206.94 MPa ~-206.94 mpa 89 Gra 89x103 mpa E- OLE &=-2.325x103 mm 1mm. =-2.325x103 mm/mm E = Elong (contracts artially) Ex Elong 9. & Ey = Elateral We know that E= 2G CI+v) V= poisson's ratio 89= 2x61) (1+v) 0.2705 89 2x61 If +0.2705 Cabsolute value)

for absolute values poisson's ratio 1v1 = • Elateral Elong 10.2705| = - Elateral 2.325x103 → Erateral ~0.629x10 mm/mm, Elaterall = 0.629x17 mm/mm. → Ad Elaterall Ad d 20mm -) 0.629x103 Ad 20 --3 ladl= 12:58x10 mm, mm; ~ 0.01258mm Therefore, Absalute change in diameter | Adl x 0.01258 mm. Adad-d d- - new diameter - 20.01258 mm