HW 6 - SECTION 5.7 Nonprismatic Beams Nonprismatic Beams...

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321 SECTION 5.7 Nonprismatic Beams Nonprismatic Beams Problem 5.7-1 A tapered cantilever beam AB of length L has square cross sections and supports a concentrated load P at the free end (see figure on the next page). The width and height of the beam vary linearly from h at the free end to h at the fixed end. A B Determine the distance x from the free end A to the cross section of maximum bending stress if h B = 3h A . What is the magnitude er max of the maximum bending stress? What is the ratio of the maximum stress to the largest stress er B at the support? Solution 5.7-1 Tapered cantilever beam P A B SQUARE CROSS SECTIONS h A = height and width at smaller end h = height and width at larger end B h, = height and width at distance x h B -=3 h A h x = h A + (h B - hA)(Z) = h A ( I + ~) S = ~ (h )3 = h~ (1 + 2x)3 x 6 x 6 L STRESS AT DISTANCE x M x 6Px erl =-= S, (h A )3( 1+ ~y AT END A: x = 0 ITA = 0 AT SUPPORT B: x = L 2PL IT =--- B 9(h A )3 CROSS SECTION OF MAXIMUM STRESS der l Set -- = 0 Evaluate the derivative, set it equal dx to zero, and solve for x. L x=- +-- 4 MAXIMUM BENDING STRESS 4PL IT max = (ITI)FLl4 = 9(h )3 A amax. --=2 +-- ITB
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332 CHAPTER 5 Stresses in Beams (Basic Topics) Solution 5.8-7 Laminated wood beam on simple supports Z in. Z in. Z in. I~ L = 6 ft = 72 in. ALLOWABLE LOAD BASED UPON BENDING STRESS Tallow = 65 psi M PL (72 in.) <Tallow = .1800 psi <T =- M = 4 = P --4- = 18P (lb-in.) S ALLOWABLE LOAD BASED UPON SHEAR STRESS 2 bh 1 . 6 . 0 4' 3 IN THE GLUED JOINTS S = --= - (41ll.)( Ill.)- = Z Ill. 6 6 VQ T=- Q = (4 in.)(Z in.)(Z in.) = 16 in. 3 (I8P lb-in.) 3P Ib <T= (P = lb; <T = psi) 24 in. 3 4 P bh 3 1 V=- 1= - = - (4 in.)(6 in.)3 = 72 in. 4 4 4 2 12 12 P 2 = - <T allow = - (1800 psi) = 2400 lb 3 3 (PI2) (16 in. 3 ) P T = = -- (P = lb; T = psi) (72in. 4 )(4in.) 36 ALLOWABLE LOAD Shear stress in the glued joints governs. PI = 36T allow = 36 (65 psi) = 2340 lb P allow = 2340 lb -+- Problem 5.8-8 A laminated plastic beam of square cross section is built up by gluing together three strips, each 10 mm X 30 mm in cross section (see figure). The beam has a total weight of 3.2 N and is simply supported with span length L = 320 mm. Considering the weight of the beam, calculate the maximum pennissib1e load P that may be placed at the midpoint if (a) the allowable shear stress in the glued J'oints is 0.3 MPa, and (b) the allowable bending stress in the plastic is 8 MPa. 30 mm ~--- L --------i Solution 5.8-8 Laminated plastic beam q h-~ bh 3 1 L = 320 rnm 4 I =- = - (30 mm) (30 mm)3 = 67 500 rnm 12 12 ' W = 3.2 N W 3.2N bh 2 1 q =-= = 10N/m S = - = - (30 mm)(30 mm)2 == 4500 mm 3 L 320 mm 6 6
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(a) ALLOWABLE LOAD BASED UPON SHEAR IN GLUED JOINTS Tallow = 0.3 MPa VQ P qL P T = - V = - + - = - + 1.6 N Ib 2 2 2 (V = newtons; P = newtons) Q = (30 mm)(lO mm)(10 mm) = 3000 mm 3 Q 3000 mm 3 I Ib (67,500 mm 4 )(30 mm) 675 mm 2 VQ P/2 + 1.6 N T--- (T=N/mm 2 =MPa) - Ib - 675 mm 2 SOLVE FOR P: P = 1350T allow - 3.2 = 405 N - 3.2 N = 402 N SECTION 5.8 Shear Stresses in Rectangular Beams
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This note was uploaded on 11/14/2011 for the course ECIV 310 taught by Professor Huckelbridge during the Spring '11 term at Case Western.

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HW 6 - SECTION 5.7 Nonprismatic Beams Nonprismatic Beams...

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