Deflection examples

Deflection examples - SEC 12.7 STATICALLY INDETERMINATE...

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Unformatted text preview: SEC. 12.7 STATICALLY INDETERMINATE BEAMS AND SHAFTS—METHOD OF INTEGRATION EXAMPEE 3248 The beam in Fig. 12—37a is fixed supported at both ends and is sub- jected to the uniform loading shown. Determine the reactions at the supports. Neglect the effect of axial load. SOEUTIGN ‘ 'i Elastic Curve. The beam deflects as shown in Fig. 12-37a. As in the (a) previous problem, only one x coordinate is necessary for the solution since the loading is continuous across the span. .Wfifiwnl Function. From the free—body diagram, Fig. 12—37b, the respective shear and moment reactions at A and B must be equal, since there is symmetry of both loading and geometry. Because of this, the equation of equilibrium, EFy = 0, requires VA 2 VB = W—L‘ :31”; 2 The beam is indeterminate to the first degree, where M’ is redundant. Using the beam segment shown in Fig. 12-37c, the internal moment M can be expressed in terms of M’ as follows: V _WL wL V _M_,_L_ A — *2- [— ——————————— —‘$ ——————————— ‘1 B _ 2 : Kigx _ g)? ‘ M, : ' ~ : ‘ V MA = M' MB = M’ finite and 3:1sz {‘mw. Applying Eq. 12—10, we have (b) dzv wL w 2 __ = __ __ ._ _._ M’ E] dx 2 x 2x VLL V1: 126—. 2 __________ __ d1) wL w i I M E—-=-—2-—3—M’ +C ,. . Idx 4x 6x x 1 x wL w M’ M I V E112 = -—-x3 - —-x4 — x2 + Clx + C2 (C) 12 24 2 he 12—37 The three unknowns, M’, C1, and C2, can be determined from the three boundary conditions 1) = 0 at x = O, which yields C2 = 0; dv/dx = 0 at x = 0, which yields C1 = 0; and '1) = 0 at x = L, which yields ' , sz , , . M 12 Am. .Using these results, notice that because of symmetry the remaining boundary condition dv/dx = O at x = L is automatically satisfied. It should be realized that this method of solution is generally suitable when only one x coordinate is needed to describe the elastic curve. If several x coordinates are needed, equations of continuity must be written, thus complicating the solution process. 630 CH. 12 DEFLECTIONS OF BEAMS AND SHAFTS EXAMT’EE 12—17 The beam is subjected to the distributed loading shown in Fig. 12—36a. Determine the reactions at A. E] is constant. 3 8 i [E T I 0 N ' 1?? Iaszic C mm: The beam deflects as shown in Fig. 12~36a. Only one coordinate x is needed. For convenience we will take it directed to the right, since the internal moment is easy to formulate. :1! mm em Firm-lion. The beam is indeterminate to the first degree as indicated from the free-body diagram, Fig. 12—36b. We can express the internal moment M in terms of the redundant force at A using the segment shown in Fig. 12—36c. Here ., \ ‘ 3 i B" M = Ayx — ’16—W0% Slope mad Eit’wtit’ Curve. Applying Eq. 12—10, we have 1 c1212 1 x3 2 E1 —- = A — fl —— poi fwdg) dxz yx 6W0 L ,’ L 4 I” ' do 1 1 x 1/4 i EI—=-A 2—— — C A ” PM dx 2 24 W L 1 2 l 5 —x*'|*-x'i V .__ _1_ 3 _ 1 L Ay 3 3 () EI’U 6Ayx 12OWOL + C1 + C2 ‘ ‘ C ‘ Fig. 12—31% The three unknowns Ay, C1, and C2 are determined from the bound— ‘ 1 ‘ ary conditions x = 0, v = 0; x = L, dv/dx = 0; and x = L, '0 = 0. Applying these conditions yields x=0,v=0; 0=OiO+O+C2 d1) 1 1 = _ = . 0 : ._ L2 - _— 3 + C ‘ Y x L, dx 0, 2A), 24W0L 1 - 1 1 x=L,v=0; 0='6"AyL3‘T2—6W0L4+C1L+C2 ‘ ‘ Solving, ‘ . 1 l 1 3 ‘3 ' Ay = EWOL r3115. t C1 = —’1—W0L3 C2 = 0 i p 120 ' Using the result for Ay, the reactions at B can be determined from the equations of equilibrium, Fig. 12—36b. Show that Bx = 0, By = 0.4w0L, and M3 = WOL2/15. CH. 12 DEFLECTIONS OF BEAMS AND SHAFTS EXAMPMZ 12—21% Determine the moment/at'B for the beam shown in Fig. 12—4911. E1 is constant. Neglect the effects of axial load. 801 U Tl @N H Principle of Superpfisizimr. Since the axial load on the beam is neglected, there will be a vertical force and moment at A and B. Here there are only two available equations of equilibrium (EM = 0, EFy = 0), and so theproblem is indeterminate to the second degree. We will assume that By and M B are redundant, so that by the prin- ciple of superposition, the beam is represented as a cantilever, loaded separately by the distributed load and reactions By and MB, Fig. 12—4917, 12—49c, and 12—4911. (b) {7111111111111311’115‘ fqumimm Referring to the displacement and I slope at B, we require ' 0+) 0 = OB + 91,9 + 9E, (1) (W) 0 = 113 + 121% + vii (2) Using the table in Appendix C to compute the slopes and dis- placements, we have wL3 _ 3 kip/ft(12 103 _ 108 (C) (d) 93 _ 48EI _ 48E] E1 ‘1, _ 7va4 7(3 kip/ft)(12 104 _ 1134 1 v3 — 384E] _ 384E] ' E1 i Only redundant M 3 applied 2 2 mg, 12-411 01,3 _ PL _ By(121”t) z 72By J 2E1 2E1 E1 , _ 11L: _ By(12 103 _ 5763, ”B_ 3EI— 3E1 — E1 i 6,, _ fl * MB(12 ft) : 12MB Q ‘ B E1 E1 E1 ,, ML2 _ MB(12 102 _ 72MB 03‘ 2E1 _ 2E1 _ E1 i Substituting these values into Eqs. 1 and 2 and canceling out the common factor E1, we get (1+) 0 = 108 + 7sz + 12MB (+1) 0 = 1134 + 576Ey + 72MB Solving these equations simultaneously gives 13y = —3.375 kip MB = 11.25 kip - 11 SEC. 12.9 STATICALLY INDETERMINATE BEAMS AND SHAFTS—METHOD OF SUPERPOSITION 645 EXAMPLE 12-22 Determine the reactions on the beam shown in Fig. 12—47a. Due to the loading and poor construction, the roller support at B settles 12 mm. Take E = 200 GPa and I = 80 (106) m4. S 0 L U TI ON Principle of Superposition. By inspection, the beam is indetermi- nate to the first degree. The roller support at B will be chosen as the redundant. The principle of superposition is shown in Fig. 12—471) and 12—47c. Here By is assumed to act upward on the beam. Compatibility qumzion. With reference to point B, using units of meters, we require (H) 0.012 m = 03 — via (1) Using the table in Appendix C, the displacements are __ 5wL4 _ 5(24 kN/rn)(8 m)4 _ 640 kN - m3 “B _ 768E] — ‘ 768EI E1 i v, _ PL3 ‘ By(8 m)3 _ 10.67 m3By B 48E] 48EI E1 T Thus Eq. 1 becomes Fig. 12—47 0.012EI = 640 - 10.678y Expressing E and I in units of kN/m2 and m4, respectively, we have 0.012(200)(106)[80(10-6)] = 640 — 10.6713y By = 42.0 kNT Ans. quilibrium Equalimzs‘. Applying this result to the beam, Fig. 12—47d, we can calculate the reactions at A and C using the equations of equilibrium. We obtain L+ EMA = 0; —96 kN(2 m) + 42.0 kN(4 m) + Cy(8 m) = 0 Cy = 3.00 kN T Ans. +T 2F}, = 0; Ay - 96 kN + 42.0 kN + 3.00 kN = 0 Ay = 51 kNT Ans. SAMPLE PROBLEM 9.8 For the uniform beam and loading shown, determine (a) the reaction at each support, (17) the slope at end A. SOLUTION Principle of Superposition. The reaction RB is designated as redundant and considered as an unknown load. The deflections due to the distributed load and to the reaction RB are considered separately as shown below. (grow (yB)lU For each loading the deflection at point B is found by using the table of Beam Deflection: and Slopes in Appendix D. . Distributed Loading. We use case 6, Appendix D w 4 3 21. + 24 [(x x3 Lx) y = — At point B, x = %L: w 2 4 2 3 WL4 : _~._ _ — 2L —L + 3 —L = —0.01132—— (WW 24EI[<3L> <3 > L 3 E1 Redundant Reaction Loading. From case 5, Appendix D, with a = %L and b = %L, we have Pazbz RB <2 >2(L)2 RBL3 = _ = + —L — = 0.01646 (M 3EIL 3EIL 3 3 E] a. Reactions at Supports. Recalling that yB = 0, we write yB z 08%; + 03);; 4 RBL3 L C 0 = —0.011321L + 0.01646 RC = 0.0413 wL E1 E1 R8 = 0.688wLT 4 Since the reaction RB is now known, we may use the methods of statics to determine the other reactions: RA = 027le T RC = 0.04l3wLT 4 b. Slope at End A. Referring again to Appendix D, we have RA = 0.271 LUL RB = 0,688 wL . . . wL3 wL3 Distributed Loading. (014)“, = :— 2 4E1 = —0.04167?1 Redundant Reaction Loading. ForP = *RB = ~0.688wL and b = %L Pb(L2 — b2) 0.688wL<L>[ (Lfl wL3 =___._~.*_=+————— L2— — 0 =0.03398-— (we 6EIL 6EIL 3 3 ( ’0" E1 Finally, 0,, = (0A)w + (9A)R L3 L3 'wL3 wL3 w w = — i — . —~ = — _ —— : 7 —— ‘ 0,, 004167 E1 + 003398 E1 000769 E] 0A 00 69 E] “N’— 563 The beam in Fig. 12—48a is fixed su connected to a §-in. 8ft 1" 5ft‘4 Actual beam and rod 5ft‘“f~“ Redundant FBC removed Only redundant FBC applied (a) (b) (C) Fig. i2—48 SOLUTION I (H) ” ~ ’ 1/ .PL Fads ft)(12 mm) 213 AB (Tr/4X2i m.)2[29(103) lap/“12] 0 01686FBC i v _ 5PL3 _ 5(8 kip)(10 ft)3(12in./ft)3 _ 01045 t B 4851 48[29(103)kip/in2](4751n4) “ ' m-i FBC(10 ft)3( 12 in./ft)3 z . 3E1 3[29(103) klp/1n2](473 m“) 004181F3CT ThUS, Eq. 1 becomes (+i) 0.01686FBC = 0.1045 - 0.04181F3C FBC 2 ...
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Deflection examples - SEC 12.7 STATICALLY INDETERMINATE...

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