{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
(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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online