Question

Question 9 1 pts w = 400 tu, - 700 #/FT 20 With the entire frame, and the equivalent concentrated loads on member BD used in Questions 7 & 8. use sum of moments at point A = 0 to determine the vertical reaction at E. Then, by drawing a free body diagram of member CE and applying the equations of equilibrium (method of members), determine the horizontal force at point D. Dx = (Hint: The force Dy determined in Question 8, and Ey, should be included on the FBD of member CE. Use sum of moments about point C = O to determine Dx) -333 lbs -449 lbs 0 -352 lbs 0 -567 lbs

Answer 1

JA 4oolbldt Troublth ما، اور او لا ENDO A (400x18x9) – (700*6* .) – 18 By=o |By = 2900(ks ] By+ Dy - (400x1&) + (700x6) Dy = 11400-2900 Dy = 8soulbs Fyzo 4ouulijat . Toofft DI Atat a y - 20 EMA=O It Ej (20) – 400X15) (10) - (100x6) (19+ :) =6 Ey = 8220 ths

| 18/ Tel . by=800) 15 - 822sys 즉 음 2 M = 0 + Xs12 Dx(Ib) + 8506(4+ 8220(0) so D - - 352lbs