02-05_chap_gere

02-05_chap_gere - 134 CHAPTER 2 Axially Loaded Numbers...

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Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a tensile stress s u 5 63 MPa and a shear stress t u 5 2 21 MPa on a certain inclined plane (see figure). Determine the stresses acting on all faces of a stress element oriented at u 5 30° and show the stresses on a sketch of the element. Solution 2.6-16 Bar in uniaxial stress 134 CHAPTER 2 Axially Loaded Numbers 21 MPa 63 MPa u s u 5 63 MPa t u 52 21 MPa I NCLINED PLANE AT ANGLE u s u 5 s x cos 2 u 63 MPa 5 s x cos 2 u (1) (2) Equate (1) and (2): or From (1) or (2): s x 5 70.0 MPa (tension) tan u 5 21 63 5 1 3 Ê u 5 18.43 8 63 MPa cos 2 u 5 21 MPa sin u cos u s x 5 21 MPa sin u cos u 2 21 MPa s x sin u cos u t u s x sin u cos u s x 5 63 MPa cos 2 u S TRESS ELEMENT AT u 5 30 8 Plane at u 5 30 81 90 85 120 8 N OTE : All stresses have units of MPa. 5 30.31 MPa t u 5 ( 2 70 MPa)(sin 120 8 )(cos 120 8 ) s u 5 (70 MPa)(cos 120 8 ) 2 5 17.5 MPa 30.31 MPa 5 ( 2 70 30 8 30 8 ) t u s x sin u cos u 5 52.5 MPa s u 5 s x cos 2 u 5 30 8 ) 2 u 30 ° 17.5 30.31 52.5 30.31 52.5 y x 0
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Problem 2.6-17 The normal stress on plane pq of a prismatic bar in tension (see figure) is found to be 7500 psi. On plane rs , which makes an angle b 5 30° with plane pq , the stress is found to be 2500 psi. Determine the maximum normal stress s max and maximum shear stress t max in the bar. Solution 2.6-17 Bar in tension SECTION 2.6 STRESSES ON INCLINED SECTIONS 135 q p r PP s b Eq. (2.29a): s u 5 s x cos 2 u b 5 30 8 P LANE pq : s 1 5 s x cos 2 u 1 s 1 5 7500 psi P LANE rs : s 2 5 s x cos 2 ( u 1 1 b ) s 2 5 2500 psi Equate s x from s 1 and s 2 : (Eq. 1) or (Eq. 2) cos 2 u 1 cos 2 ( u 1 1 b ) 5 s 1 s 2 Ê cos u 1 cos( u 1 1 b ) 5 B s 1 s 2 s x 5 s 1 cos 2 u 1 5 s 2 cos 2 ( u 1 1 b ) S UBSTITUTE NUMERICAL VALUES INTO E Q . (2): Solve by iteration or a computer program: u 1 5 30 8 M AXIMUM NORMAL STRESS ( FROM E Q . 1) M AXIMUM SHEAR STRESS t max 5 s x 2 5 5,000 psi 5 10,000 psi s max 5 s x 5 s 1 cos 2 u 1 5 7500 psi cos 2 30 8 cos u 1 cos( u 1 1 30 8 ) 5 B 7500 psi 2500 psi 5 Ï 3 5 1.7321 q p r s b
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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 u 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. (a) Determine the angle u 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 P max if the cross-sectional area of the bar is 225 mm 2 . Solution 2.6-18 Bar in tension with glued joint 136 CHAPTER 2 Axially Loaded Numbers q p PP u 25 8, u , 45 8 A 5 225 mm 2 On glued joint: s allow 5 5.0 MPa t allow 5 3.0 MPa A LLOWABLE STRESS s x IN TENSION (1) t u 52 s x sin u cos u Since the direction of t u is immaterial, we can write: | t u | 5 s x sin u cos u or (2) G RAPH OF E QS . (1) AND (2) s x 5 Z t u Z sin u cos u 5 3.0 MPa sin u cos u s u 5 s x cos 2 u Ê s x 5 s u cos 2 u 5 5.0 MPa cos 2 u (a) D ETERMINE ANGLE U FOR LARGEST LOAD Point A gives the largest value of s x and hence the largest load. To determine the angle u corresponding to point A , we equate Eqs. (1) and (2).
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This note was uploaded on 04/30/2008 for the course CVEN 305 taught by Professor Gardoni during the Spring '08 term at Texas A&M.

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02-05_chap_gere - 134 CHAPTER 2 Axially Loaded Numbers...

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