Question

(a) A cube of concrete 100 mm on each edge is compressed in two perpendicular directions by forces P = 67 kN. Determine the change AV in the volume of the cube and the total strain energy U stored in the cube assuming E = 26 GPa and V = 0.1. (b) For the given state of stress shown (Fig. Q1b), determine: (i) The principal stresses. (ii) The corresponding principal planes. 48 MPa 16 MPa 60 MPa Fig. 21(b) Page 2 of 3

Answer 1

ASSUMPTIONS :

1.) Stress is uniform

2.) Material is homogenous

3.) Material is Isotropic

a.) Given Side of Cube (a) LOOmm 0.1m Compressive force (P) = +67 KN - - 67 KN Modulus of elasticity (E) - 26 Gla Poission's Ratio () = 0.1 67 KN We Know that, * 1-28 67% Δν Vi Ontry) E יו pi 3 -67x103 (0.1) z -6700 x 10°N 2 2 m2 AV 2(0:1) 26 X109 (0.113 3 6700 x 10 Fax ot : AV =- 4.123 x107 m3 (or) AV -412.307 mm * 3 : . Strain energy stored (V)- lbh [xfnt og ty 2 where, En น pl E where, Trang Goy on a von :. Ent Ey = 2(1-0) (..) a

:: Uz l3 2 on (Ext cy) as on 2 (1-1) on 1.03 09 (1-1) ) Uz (0.19 (1-0-1) 6700x103) E 2 26 x 109 * U= 1.553 Joules b.) Given Stress element is, 48 MPa را In = 16 MPa Ty :-48 MPa Try -60 MPa we know thah Principal stresses are, toy 1,2 16 MPa 60 MPa 2 + Tuy 2 (****) () 0,2 = 116 -48 + (16 +48) ? + (60)2 2 ) 12 -16 + 68 어 -16 + 68 an -16-68 * .., 어 52 MPa 11 :lo =-84 MPa 2

I The Principal Planes are, Tan-1/2 Try (1 Principal plane) Opa Op, + 90° (and principal plane) - Opi info Tant 26-60) 16 + 48 0 lop, -30.963° * OP -30.963 + 90 Op = 59.03 THE END X