Question

Learning Goal: To solve for forces in statically indeterminate bars with axial loads. Part A-Force with a known deflection When the number of reaction forces is greater than the number of equilibrium equations, the system is statically indeterminate. Solving for the reactions requires some additional equations. These additional equations come from considering compatibility relationships (i.e., continuity of displacements and relationships between displacements and loads). The square bar shown (Figure 1) is 72.5 mm thick and 3.6 m long and is fixed supported at both ends. A load directed leftward is applied at point C, as shown, L2 = 2.74 m from the left end. The modulus of elasticity is E = 100 GPa If point C moves 8 = 0.200 mm to the left, what is the applied force? Express your answer with appropriate units to three significant figures. View Available Hint(s) UA ? For an axially loaded member, the compatibility relationship for the deflections can be written by setting the total relative axial displacement between the ends of the member to a known value. NL The load-displacement relationship á = Σ AE gives another equation for the deflections. Once F= 226.08 KN Submit Previous Answers igure < 1 of 2 X Incorrect; Try Again; One attempt remaining C B Part B - Reaction with a known load F L2 Consider a new structure, where the thickness of the bar is reduced to 36.25 mm from C to B (it is still square) (Figure 2) and I = 2.74 m. If the applied load is F = 390 kN ,then what is the reaction at A? Let a positive reaction act to the right. The total length is still 3.6 m Express your answer with appropriate units to three significant figures.

Answer 1

Given that Square bar of thic keren (t) = 72.5mm Total length of bar (L) = 36 m Moderkes of flarticiley (E) E 100 G Pa Part-A!- We know that, §= FL AE since the force applied at C, couie expansion in BC and contraction in AC ie Eupanion in BC = contraction in Ac. (be) BC Odbaci (5) BC = (slac Let Lot F be a part pardt force in " used for eupanien in Bc ,lie) F, ause eupanion in BC-Vod' and Fq be the force causes compression in "AC" SO, Total Force E - F + F₂ from 0, implies FIL, FALL

. Fili = q la Fx (3.6–2.74) - F{X2.74 Fix 0.86 {12.74 f, ~ 3. 186*FL ALO {= 6 = {le AE =8 » F = AES & = 0.072580.0725x 100 x 10 xo 700 xro” x 0.2 rro .: 0.74 r = 38.36667 KR so, q ~ 38.3667 * 3.186 F = 122-2365 kPa since F = E +Fq from ③ F = 122.236 + 38-3667 F160.60L KRT

Part-B: cic B А from 0 RA RR Fill AIE Fil AgE Now [AR #AL] Fix2.74 مراد : F2X0.86 d = 0.86 d=2.74 0.03625 KO. 03618 8:0725*0.0725 iilF 2 12.74417, from 9 - Fit F = F 12.70417,+f, Et us = 390 28.345 keel 13.7041 Since EF 4 = 0 on AC. SO, RA-f=0 af = 28.375 KN so, Reaction af A (P=28.375 kn

Part-Ci- Equal reactions of a hea Fiah .(RAR) sa from 0 a 8 = 82 Fill Fil A, E ARE .:[h=) (-) R-u AL AI - 3.6-a 0.0347X0.03625 0.0725 YO 0725. 4.9763 = 3-6 36 +1 ? lle 3.6 0.60237 m 5.476) Therefore in should be = 0.6027)
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