# HW8 - 12.27 The stresses shown in the figure act at a point...

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12.27 The stresses shown in the figure act at a point in a stressed body. Determine the normal and shear stresses at this point on the inclined plane shown. Fig. P12.27 Solution The given stress values are: 5.5 ksi, 18.7 ksi, 0 ksi, 20 x y xy     The normal stress transformation equation [Eq. (12.3)] gives n : 22 cos sin 2 sin cos ( 5.5 ksi)cos ( 20 ) (18.7 ksi)sin ( 20 ) 2(0 ksi)sin( 20 )cos( 20 2.67 k ) 2.6691 ksi si (C) n x y xy         Ans. The shear stress transformation equation [Eq. (12.4)] gives nt : ( )sin cos (cos sin ) [( 5.5 ksi) (18.7 ksi)]sin( 20 )cos( 20 ) (0 ksi)[cos ( 20 ) sin ( 20 )] 7.7777 k 7.7 ks s 8 i i nt x y xy           Ans.

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12.37 The stresses shown in Fig. P12.37 a act at a point on the free surface of a stressed body. Determine the normal stresses n and t and the shear stress nt at this point if they act on the rotated stress element shown in Fig. P12.37 b . ( a ) ( b ) Fig. P12.37 Solution The given stress values are: 50 MPa, 15 MPa, 40 MPa, 36 x y xy       The normal stress transformation equation [Eq. (12.3)] gives n : 22 cos sin 2 sin cos (50 MPa)cos ( 36 ) ( 15 MPa)sin ( 36 ) 2( 40 MPa)sin( 36 )cos( 36 ) 65.5 65.6 MPa ( 853 MP T) a n x y xy       Ans. To find t , add 90° to the value of used in Eq. (12.3): cos sin 2 sin cos (50 MPa)cos ( 36 90 ) ( 15 MPa)sin ( 36 90 ) 2( 40 MPa)sin( 36 90 )cos( 36 90 ) (50 MPa)cos (54 ) ( 15 MPa)sin (54 ) 2( 40 MPa)sin(54 )cos(54 ) 30. 30.6 5853 M a MP P t x y xy                     a (C) Ans. The shear stress transformation equation [Eq. (12.4)] gives nt : ( )sin cos (cos sin ) [(50 MPa) ( 15 MPa)]sin( 36 )cos( 36 ) ( 40 MPa)[cos ( 36 ) sin ( 36 )] 18.5487 MP 18.5 M a 5 Pa nt x y xy            Ans.
12.41 Consider a point in a structural member that is subjected to plane stress.

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HW8 - 12.27 The stresses shown in the figure act at a point...

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