HW9 - PROBLEM 2.79 2.79 The plastic block shown 18 bonded...

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Unformatted text preview: PROBLEM 2.79 2.79 The plastic block shown 18 bonded to a fixed base and to a honzontal I'lgld plate to which a force Pis applied. Knowing that for the plastic used G = 55 ksi, determine the deflection of the plate when P = 9 kips. ? Cansr’olev‘ ‘Hle papas‘i'ic bpock. The shear-Wu: Force Carrie) is P = 9:1031L‘ ‘——1> A = (3.5)(5.5)= 19.25 'm‘ 467. 52 PSI. Sheam‘ma 3+V‘aim 7‘ 1' “é— : %% 7 0. 0085006 is 'V = (2.2 3010085006) = 0.0187 in, 2.80 A vibration isolation unit consists of two blocks of hard rubber bonded to plate AB and to rigid supports as shown. For the type and grade of rubber used 2;“ = 220 psi and G = 1800 psi. Knowing that a centric vertical force of magnitude P; 3.2 kips must cause a 0.1 in. vertical deflection of the plate AB, determine the smallest allowable dimensions a and b of the block. SOLUTION Consiaicr +Le router Mode on Hm N3 H. I+ Carries a. shearing 4’0»ch equal +9 ~513— Tlde sheqn‘nj 5+ress is f = a2 3 OV‘ required A = -E -——3‘2x‘o : 12727 in“ 22’ - (23(220) 130+ A = (3.0) b - A - - Hence 5 - 3.0 - 2.42 m. "-1 USE b? 1‘”! ha and 2": 220 Pfit' Shea-ring S+Nt€n Y 7 g = 1:3): = 0.12222. 11> * B + 7” = 5 #1 l " E HCI’ICQ PROBLEM 2.81 PROBLEM 2.82 2.81 An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 10 mm when a 22 kN lateral load. is applied as shown. Determine (a) the smallest allowable dimension b, (b) the smallest required thickness 0 if the maximum allowable shearing stress is 420 kPa. SOLUTION 5‘1 Banks $0.112 P T 227103 N Shear-M») stress 7: “ [+20 VIC:3 Pa -3 . .:_E__ axto‘ _ -3 1. ’Zf- A on A t-fim o52.38lX/O M .— 52.38! x103 m...‘ A 1‘ (200 MM)( .38 3 - 52 1x10 2 262 Mm ‘+é€.é7 x163 2I.Ll mm 2.82 For the elastomeric bearing in Prob. 2.81 with b = 220 mm and a = 30 mm, determine the shearing modulus G and the shear stress rfor a maximum lateral load P = 19 kN and a maximum displacement 6= 12 mm. SOLUTION Shearivfl ¥ome P = l‘? x103 N Ana A 1' (ZOO Mh\ ? H‘IXIO: Min-L = Wino“ In" NH}: _ mxlo3 ‘ c. _ ~ 431.8lxtospa A qq’dos = 43! kPa Shequnfl si’rafn v: % = —‘3;:—:: ’—'- 0.400 Shearu‘n 3 MOJJIUS N 3 92%.: W = Loaoxlo‘ Pa = L080 M95. 4 ...
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HW9 - PROBLEM 2.79 2.79 The plastic block shown 18 bonded...

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