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Hw3 - Bars 1 The bar is fixed at ends A and point B and...

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Unformatted text preview: Bars: 1. The bar is fixed at ends A and point B and force P = 50 kN is acting at C. The area ofthe cross section is 200 mm2 and the modulus of elasticity E is 200 GPa. Find the horizontal reactions at A and B. ,4 E C 4 <—'—>1m |<—4m—>t<—1m_p| Equilibrium XFX=0 a SOkN—FA+FB=0 FA —FB =50kN olAA—ord, =50kN A=AA =A, =200mm2 g _50,000N 0‘ —o, 00 mm: = 250 MPa (i) v . . V . P L (P —P)L Compatibility Condition 6””, = 6A —6B = 0 = QE" 4.? 5A =_65 8.4144 =_£BLB —L = L = 0' =-0' — E A E B A 8 LA Substitute back from (i) O" — O'B = 250 MPa L —o,, L—‘ —a, = 250 MPa A 0' B = —200 M Pa (compression) 0" = 50 MPa (tension) FA = O'AAA = 50 MPa* 200 mm2 =10 kN (tension) F, = 03A, = 200 MPa" 200 mmz = 40 W (compression) Bars: 2. A steel bolt and nut are tightened around an aluminum sleeve. Given the following cross sectional areas (As-r & AAL). thermal expansion coefficients (as-[- & aAL) and moduli of elasticity (E51- & EAL) for both the steel and aluminum. find the stress in the steel (051) and the stress in the aluminum (GAL) if a change in temperame (A1) of 65°C is imposed. x \\\\\\\\\V I m L \\\\\\\\\s I Initial Position A31- : 400 mm2 AAL = 600 mm: asr=12x1041°c (m: 23x 10‘6/°C E5T=200 GPa 51:73.1 GPa AI = 65 °c Solution .' We know: F51. = FM (i) Compatibility Condition: 6,0,“ = (5M )1 — (5“ )F = (537. )I. + (53,. )1, F X0.150 F X0.150 23x]0"’ x65 °Cx0.150 m—“fifi—mg =12x10‘° x65 °C><0.l50 m+% (ii) 600x10 X73.leO 400x10 X200X10 Solve equations (i) and (ii): FST = F“ = 20.26 kN 20260 N V 20260 0' =—=50.6 MP 0' =—=33.8MP ( 57' )bn!: 1 0—6 a ( Al. )xlww 10—6 0 Bars: 3. A steel bolt and nut are tightened around an aluminum sleeve. Find the internal forces as the nut is rotated l tum (1 turn = 1.5 mm) given the cross sectional areas and the moduli of elasticity for the steel and the aluminum. V V V Agr=150mm2AAL=100mm2 a a a L=150 mm E5T=200 GPa EAL: 70 GPa / y a i I 9 . a a 5 Fm; I f i 1.5 mm Posmon A A A AAL From statics: F5, = FM = F salutiom Compatibility: A5, + A”, = 1.5 mm Fixms 9+ [326x015 fillsmlo 150x10 x200x10 100x10 x70x10 00 5x10‘° [-‘+2.15x10'8 F=1.5 F=ixlo5 =56.6 kN 2.65 I 000 _ 56600 N AST = 377.4 MPa _ 56600 N AA]. = 566.6 MPa Bars: 5. For the, find the total displacement of point A (5A). The bar has a modulus of elasticity E=200 GPa. 600 mm A=600 mm2 400 mm A=200 mm2 P=200 kN Solution L 6 = J P’dx (General) 0 AXE mo 400 6=I xdx+ I P,dx 0 AB 0 AH 6 = PL“. + PL" AME AME _ 200x103 x600x10'3 + 200x103 x400x10‘3 6' —6 9 —6 9 600x10 x200x10 200x10 x200x10 5=lmm+2mm Bars: 15 A rigid bar OBC of length L is supported at three points as shown. It is pinned at O and connected to two elastic bars BD and CF. BD and CF have pin joints at both ends, are made of a material with elastic modulus of E and are of length 5. The area of cross section of BD is A and that of CF is 2A. A force P is applied at C as shown. (a) Determine the downward deflection 8 of point C (in terms of P, E, A and 5). What are the forces in BD and CF in terms of P? (b) The temperature of BD and CF is then increased by AT. The coefficient of thermal expansion of BD and CF is a. What are the forces in BD and CF due to the combined effect of the load P and the temperature increase AT (in terms of P, E, A, AT and a)? Note that the temperature increase has no effect on the rigid bar. (c) At What temperature AT (in terms of P, E, A and a) does the force in BD become zero? Solution (3) Let F39, F02 denote the force in bars BD and CF, respectively. Moment equilibrium about 0 implies that L l (P—FCF)L = FEDS —)(P—FC,,) = F503 . . . . . 1 F s 1 F s l Compatibility implies that 65D = 55”, —) T3 = 5 5A —) FED = ZFCF The above 2 equations yield F RD = g P, FCF =31) 4 PS Th , 6 = —— e" CF 9 EA (b) Now the forces in BD and CF will change. Elongation of bar BD: 51m = % + a (AT)S - FCFS Elongation of bar CF: 50.. = EA +0: (AT)S Compatibility: 263,, = 60,.- L l Moment equilibrium about 0: (P - FCF )L = F RD 3 -) (P - FCF) = FED E Compatibility and moment equilibrium yield 2 equations for the 2 unknown forces: FBD=%P—%AEa(AT), FC =§p+§AEa(AT) Bars: 18 Find the total elongation of the rod system shown hanging under its own weight with cross-sectional areas, A 1 and A 2, weight density, 7, and elastic modulus, E. Solution For the rod of length L2, using a coordinate system origin at its bottom, the force at distance P“) = yA‘x . Then, if 52 denotes the elongation of the rod of length L2, xis “P L* 1 0 0 For the rod of length L1, using a coordinate system positioned at its bottom (0 S x S Ll) P(x) = 7A1“ 7x1sz Then, if 51 denotes the elongation of the rod of length 12, 54PM”? Emu: Fl; 2+yA2L2I1 '0A,E 0 E A13 215 AIE _71 2 2 A2 62—E[§(L1+L2)+ZL1L2) 00 || 30 + ...
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