Question

C2 Done session.masteringengine AA <HW11 - Attempt 1 Problem 4 < 4 of 4 The frame is subjected to the loading shown in. 1sc = 800 in -12 ft 8 ft P 10=550 in 8 ft Assume the joint at B is fixed, C is pinned, and A is fixed. The moment of inertia of each member is listed in the figure. Suppose that E = 29(103) ksi, P=6 k, w = 2.2 k/ft. Part A

Answer 1

given We 2.2klft P=6k & LAB= IBC = gooing OFE = 500 iny 8Ft E- 29000lesi Ant * 12Ft X fixed and moment MAB= - 6x16 = 12 kft MBA 12 kft +6x16 MBC= - 2. 2X122 39.6 kft 8 mon 39.6 kft thelle is (No sway) A e floke deflection egh m MAB -12 + 9X99000X580 (204 +03) XL 16 122 MBA +12 + 2x 29000X 590 16 908464) X1

MCB = 439.6 + 8x29000 y 6oo. 2 OB TO c) x 1 122 MBC= — 39.6 + px 39000 X800 12 (202 + )** 12 12 Joint equillibrium OA=0 fixed. (A) MBA & MB C =0 ( 28846)= Het Mpg 000 XSS 0 / 90g) + -39.6 + 9x 29000 XBOO 76x12² 12x12² H2 + 138456, 486 (206) — 39.6 +2685).85 $(206+04)=0 ☺ ☺ 81394.675708 + 86851.850C = 87.6 MCB=0 +39.6 + 2429000X800 ( 20e + OB) = 0 12x12² 236851.85185 08 + 53703.7037-39.6

De = -1.086X163 31.33 kft macz -39,6+ X 29000 4800 from egh A & B OB = 6.978104 end momenty fart A MAB= +2+ 2X29000XSSO 6.97804) --23497kft 16X122 Parts MBA = (t exgoooysso 286.97x109) 16x122 Parte Parto MEB = +39.6 + 242900 X 80 0 | 2x-1886 xro? + G9rxiol) finally Parta farte MAB 2.35 kft MACE – 31.33 kft Paute Party MBA = 31.33 kft MCB= ma 123 123 0