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193_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 193_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 187 187 SECTION 2.6 Stresses on Inclined Sections L 18 mm — 2 and h 40 mm) is held snugly (but without any initial stress) between rigid supports (see figure). The allowable stresses on the inclined plane pq bL at midspan, for which 55°, are specified as 60 MPa in compression p and 30 MPa in shear. P (a) What is the maximum permissible temperature rise T if the allowable stresses on plane pq are not to be exceeded? (Assume A 17 10 6/°C and E 120 GPa.) Load for part (c) only (b) If the temperature increases by the maximum permissible amount, what are the stresses on plane pq? (c) If the temperature rise T 28°C, how far to the right of end A (distance L, expressed as a fraction of length L) can load P 15 kN be applied without exceeding allowable stress values in the bar? Assume that a 75 MPa and a 35 MPa. L — 2 Problem 2.6-12 A copper bar of rectangular cross section (b b u h B q Solution 2.6-12 (b) STRESSES ON PLANE PQ FOR max. TEMP. E x xcos( pq p 55 a b 180 u b A T radians pq s x1 s x2 sx E Tmax ¢ Tmax Ea 2 xcos( ) pq xsin( ) cos( ) ^ set each equal to corresponding allowable & solve for x s pqa 182.38 MPa x1 cos1u22 t pqa sin1u2cos1u2 s x2 Ea ) cos( ) pq 30 MPa ; ; 28 DEGREES C P 15 kN from one-degree stat-indet analysis, reactions RA & RB due to load P are: x2 63.85 MPa lesser value controls so allowable shear stress governs ¢Tmax 21.0 MPa pq T (a) FIND Tmax BASED ON ALLOWABLE NORMAL & SHEAR STRESS VALUES ON PLANE pq x 63.85 MPa x (c) ADD LOAD P IN X-DIRECTION TO TEMPERATURE CHANGE & FIND LOCATION OF LOAD 18 mm h 40 mm bh A 720 mm2 60 MPa 30 Mpa pqa 6 17 (10 )/°C E 120 GPa 20°C P 15 kN pqa )2 xsin( pq NUMERICAL DATA Tmax Tmax 31.3°C ; RA (1 )P RB P now add normal stresses due to P to thermal stresses due to T (tension in segment 0 to L, compression in segment L to L) Stresses in bar (0 to L) sx Ea ¢ T + shear controls so set 2t a b E a ¢T + 1 RA A sx 2 a & solve for tmax max b )P (1 A A [2 t a + Ea ¢ T] P 5.1 ^ impossible so evaluate segment ( L to L) ...
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