_______________________________________________________________________ PROBLEM (8.15) At a point in a loaded structure, the normal and shear stresses have the magnitudes and directions acting on the inclined element depicted in Fig. P8.15. Calculate the stresses τ xy σx , σy , on an element whose sides are parallel to the xy axes. SOLUTION Transform from θ = 30o to θ = 0 . For convenience in computations, Let σ x = −200 MPa, σ y = −60 MPa, τ xy = 35 MPa and θ = −30o Then 1 1 2 2 1 1 = (−200 − 60) + (−200 + 60) cos(−60o ) + 35sin(−60o ) 2 2 = −195.3 MPa σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ 1 2 1 = − (−200 + 60) sin(−60o ) + 35cos(−60o ) 2 = −43.1 MPa τ x ' y ' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ So σ y ' = σ x + σ y − σ x ' = −200 − 60 + 195.3 = −64.7 MPa For θ = 0o : y 64.7 MPa 43.1 MPa 195.3 MPa x Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. and