199_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

199_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.6.qxd 9/25/08 11:40 AM Page 193 193 SECTION 2.6 Stresses on Inclined Sections Solution 2.6-17 Bar in tension 23 A 2500 psi SUBSTITUTE NUMERICAL VALUES INTO EQ. (2): cosu1 7500 psi ) cos(u1 + 30° 1.7321 Solve by iteration or a computer program: 30° 1 Eq. (2-29a): MAXIMUM NORMAL STRESS (FROM EQ. 1) 2 xcos 30° smax PLANE pq: PLANE rs: Equate x 1 2 from 2 xcos 1 2 xcos ( 1 1 and ) 7500 psi 2 2500 psi s2 cos2u1 cos2(u1 + b ) or cos2u1 2 cos (u1 + b ) cosu1 s1 s2 cos(u1 + b ) s1 7500 psi 2 cos2 30° cos u1 ; 10,000 psi MAXIMUM SHEAR STRESS 2: s1 sx 1 sx (Eq. 1) A s2 s1 tmax sx 2 5,000 psi ; (Eq. 2) Problem 2.6-18 A tension member is to be constructed of two pieces of plastic glued along plane pq (see figure). For purposes of cutting and gluing, the angle must be between 25° and 45°. The allowable stresses on the glued joint in tension and shear are 5.0 MPa and 3.0 MPa, respectively. u p P P q (a) Determine the angle so that the bar will carry the largest load P. (Assume that the strength of the glued joint controls the design.) (b) Determine the maximum allowable load Pmax if the cross-sectional area of the bar is 225 mm2. Solution 2.6-18 Bar in tension with glued joint ALLOWABLE STRESS 2 xcos 25° A xsin 45° sx allow 5.0 MPa allow 3.0 MPa su 2 cos u 5.0 MPa cos2u (1) cos Since the direction of | cos xsin 225 mm2 On glued joint: x IN TENSION is immaterial, we can write: ...
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