Axially loaded

Axially loaded - ways re in— given were f the new)rob—...

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Unformatted text preview: ways re in— given were f the new )rob— at be :and :qua- ; ob— cs is rces, : fol- es in if all rams 2.11) 3 un- y in— e de— Re“ 2.12) 1.13) ,1 for em- "iAMPLE. 2-03 a ' A bar AB of length L and uniform cross section is attached to rigid supports at A and B before being loaded. What are the stresses in portions AC and BC due to the application of a load P at point C (Fig. 2.2661)? Drawing the free-body diagram of the bar (Fig. 2.2617). we obtain the equilibrium equation RA + RB = P (2.14) Since this equation is not sufficient to determine the two un- known reactions RA and R3, the problem is statically indeter- minate. However, the reactions may be determined if we observe from the geometry that the total elongation 5 of the bar must be zero. Denoting by 61 and 52, respectively, the elongations of the portions AC and BC , we write 5=5l+52=0 or, expressing 51 and 82 in terms of the corresponding internal forces PI and P2: 5—P‘L1+P2L3—0 715 _ AE AE _ (“‘ ) But we note from the free—body diagrams shown respectively in parts b and c of Fig. 2.27 that P1 = RA and P2 = —R3. Car« rying these values into (2.15), we write RALI ~ RBLZ = 0 (2.16) Equations (2.14) and (2.16) can be solved simultaneously for RA and RB; we obtain RA = PIC/L and RB = PLl/L. The desired stresses 0'l in AC and 02 in BC are obtained by dividing, respectively, P1 = RA and P2 = —RB by the cross~ sectional area of the bar: PL2 PLl = W. a? _ _ AL - AL (71 M Superposition Method. We observe that a structure is statically in— determinate whenever it is held by more supports than are required to maintain its equilibrium. This results in more unknown reactions than available equilibrium equations. It is often found convenient to desig— nate one of the reactions as redundant and to eliminate the correspon— ding support. Since the stated conditions of the problem cannot be ar— bitrarily changed, the redundant reaction must be maintained in the solution. But it will be treated as an unknown load which, together with the other loads, must produce deformations that are compatible with the original constraints. The actual solution of the problem is carried out by considering separately the deformations caused by the given loads and by the redundant reaction, and by adding—or superposing~the re— sults obtainedt i‘The general conditions under which the combined effect of several loads can be obtained in this way are discussed in Sec. 2.12. 1‘ ' umme-mwwmiwwu SEC. 4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 139 p, ,. ”= g,“ fifi’ii “be “r The steel rod shown in Fig. 4—1251 has a diameter of 5 mm. It is attached to the fixed wall at A, and before it is loaded there is a gap between the wall at B’ and the rod of 1 mm. Determine the reactions at A and B’ if the rod is subjected to an axial force of P = 20 kN as shown. Neglect the size of the collar at C. Take Est = 200 GPa. As shown on the free—body diagram, Fig. 4—12b, we will assume that the force P is large enough to cause the rod’s end B to contact the wall at B’. The problem is statically indeterminate since FA there are two unknowns and only one equation of equilibrium. ’ Equilibrium of the rod requires 23 2F, = 0; —FA — FB + 20(103) N = 0 ' " The compatibility condition for the rod is 53/A = 0.001 m This displacement can be expressed in terms of the unknown reactions by using the load—displacement relationship, Eq. 4—2, applied to segments AC and CB, Fig. 4~12c. Working in units of newtons and meters, we have _ _ FALAC _ FBLCB 5B/A — In —- AE AE w m) 0001 m _ 710.0025 m)2[200(109) N/mz] FB(O.8 m) 710.0025 m)2[200(109) N/mz] FA(O.4 m) — FB(O.8 m) = 3927.0 N ~ in (2) Solving Eqs. 1 and 2 yields FA = 16.6 kN FB = 3.40 kN Since the answer for F B is positive, indeed the end B contacts the wall at B’ as originally assumed. On the other hand, if F B were a negative quantity, the problem would be statically determinate, so that F B = 0 and F A = 20 kN. Ehi—Ehiifii §~7 The three A-36 steel bars shown in Fig. 4-14a are pin—connected to a rigid member. If the applied load on the member is 15 kN, deter— mine the force developed in each bar. Bars AB and EF each have a cross—sectional area of 25 m2, and bar CD has a cross-sectional area of 15 m2. ESL? Eiiii‘é ifn‘igmz. The free-body diagram of the rigid member is shown in Fig. 4—14b. This problem is statically indeterminate since there are three unknowns and only two available equilibrium equations. These equations are +1‘ 2Fy = 0; FA + FC + FE— 15 kN = 0 (1) 1+ ZMC = 0; —FA(O.4 m) + 15 kN(0.2 m) + FE(0.4 m) = 0 (2) if ‘umpmibiiiz}: Due to the displacements at the ends of each bar, line ACE shown in Fig. 4—14c will move to the inclined points A’C’E’. From this position, the displacements of points A, C, and E can be related by proportional triangles. Thus the compatibility equation for these displacements is 5A_5E_ 5C—5E 0.8m “ 0.4m 1 1 5=—5 +——a C 2A 2E Using the load—displacement relationship, Eq. 4—2, we have FCL 1[ PAL ] 1[ FEL'] = + (15 mm2)Est 2 (25 mm2)Est 2 FC = 0.3FA ‘l‘ 0.3FE (25 mm2)Est Solving Eqs. 1—3 simultaneously yields FA = 9.52 kN FC = 3.46 kN FE = 2.02 kN i8 kN/m Probs. 4—64/4—65 *4—64. The horizontal beam is assumed to be rigid and suppons the distributed load shown. Determine the vertical reactions at the supports. Each support consists of a wooden post having a diame- ter of 120 mm and an unloaded (original) length of 1.40 m. Take EN = 12 GPn. _..__——-—._.—....— \ SC’OCA' j \ \ \‘p w; L .2ng '- OCB’EK _ + AE ' .'.. .. w.aumnu.gmsu_.u+. .'... “Lungs a r—————12in-—~—“I SAMPLE PROBLEM The rigid casting DEF is held by two gin—diameter bolts AD and CF against the 1.5—in.-diameter rod BE. The pitch of each bolt thread is 0.]. in.. and after being snugly fitted. each nut is tightened one quarter of a turn. Knewing that for steel E = 30,000 ksi and for aluminum E = 10.000 ksi. determine the normal stress in the rod. r'AIuminumf; 2" Statics. Considering the free body of the entire assembly, we note that the R . -l \ reactions are statically indeterminate. However. We have by symmetry RA = R0, R B and using statics we write '3 RF C El i SF = 0: —RA + RS _ RC = 0 or RB = 233 (l) Deformations. We shall use the method of superposition and consider the . reaction RB as redundant. The connection at B is temporarily removed and, as the nuts are tightened one quarter of a turn. the casting DEF and rod BE move to » the left (Fig. l) a distance 51 =1-(0.lin.) = 0.025 in. 1—— The reaction R3 is then applied to force the end B to move back to the right through 8.2 (Fig. 2). Since the final deflection 6 of B must be zero \Fig. 3), we have 7 82 = 51. Noting that the deflection 82 is the sum of the deformations of bolt AD and rod BE, we write ' R L R 1" . 8mm : 2 : L132 X 10—33; ‘ film.) (30 x 1031611 R R 12‘ . 8m = if = , 3‘ m) = 0.679 x roe-3123 ‘ $1.5 in.)2(10 X 10-5 ksil abolt ‘l' Srod = ‘5: = ‘51 1.132 X 10—3111i + 0.679 X 10—333 = 0.025 in. (2) Reactions. Substituting for 8,, from i into (2): V 1.132 X 10‘3RA + 0.679 iD‘TiRA) = 0.025 in. RA = 10.04 kips RB = 234 = 2(10.04 laps) = 20.08 hips Stress in Rod ...
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Axially loaded - ways re in— given were f the new)rob—...

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