Question

QUESTION 1 In the frame shown below dimension His 7m, y is 3m, L is 7m, x is 4m, I is 4m, h is 3m. External loads are: the horizontal point load P of 2kN is applied at B, the clockwise moment M of 5kNm is applied at D, the vertical uniformly distributed load w of 1kN/m is applied on bar FJ. Calculate reactions of supports. Note that this frame has two internal pins; at Cand at F. What is the vertical reaction at pin support G? Give you answer in kN with 2 decimal points accuracy. Don't write units * L II QUESTION 2 One systematic way to determine internal actions in a 2D frame is to divide or "brake" it into a number of simpler free bodies (usually but not necessarily bars) and consider equilibrium of these free bodies one at a time. Internal forces and moments are calculated progressively for each free body and transferred onto the next one. Which of the following sequences of free bodies can be used in the written order for the analysis of the 2D frame from previous question? JK FJ, ABC CDE EFG ABC FJK, EFG, CDE CDE, EFG, ABC FJ, JK FG, EF, FJ,JK DE, CD, BC, AB

Answer 1

From ABC, Mc=0 Ax(7) ==2(4) M5 KN. Ik Ax=-11143KN . 1.143 KN(K) from FJK, P-2kN K EME=0 Ky (4) = 1(4) (4) - Ky = 2KN ]60) Ax L 4m x Ayk 7m x Тky EF=0 7 AxtGx + 20 * | * 4m) 3 -1.143+Gu+2=0 = G2 = -0.857kn) IM-o From (Eight side ) ie CoE & EFG Gy[7) + ky (11) + Gx (7) = 1(4)[7+ų) +5 Gy (7) +2011) -0.857 (7) = 41. Vertical Reaction ad os sy = 3-6846 kN (1) Gy = 3.5712 KN Ify =0 > Ay + Gytky = 3 (1) = 24 Ay +3.5812 +2 =245) Aqee3 Ay =-1.5712 KN | Ion Ay= 1.5712KN (1) ① Free Body Free Body diagrams are considered as ABC, FJK, EFG and CDE. with the help these, we are calculated Reactions.