Academic Year : 2016-2017 Semester : First Level : Second Time : 1½ Hours Date : 22 / 11 / 2016 Examiner: Dr. M. Abdel-Kader Ministry of Higher Education Giza Higher Institute for Eng. & Tech. Civil Engineering Department Course Name: Theory of Structures (1) Course Code : CIV 201 Answer of Mid-Term Exam 34 kN.m Question (1): (9 Marks) For the shown structure, determine the reactions at the supports A, B and C and the force in the link member AE. 28.5 kN 8 kN/m 4 kN/m B H C E D 3m A 4m 4 4 4 3m 28.5 kN Part DC: + M D 0 : 8 kN/m (8×4)(2) + (28.5)(4) + (½×8×3)(5) – Cy(7) = 0 + Fy 0 : D Cy = 34 kN Dx Dy + Cy – 32 – 28.5 – 12 = 0 Part HD C Cx Dy 4 Dy = 38.5 kN 4 kN/m D B H E Dx 0.6FAE 4 FAE 4m By 4 –34 – (4×4)(6) – (0.6FAE)(4) + (38.5)(4) = 0 FAE = +10 kN + Fy 0 : Cy 38.5 kN 34 kN.m + M B 0 : 3m FAE = 10 kN –16 – 0.6(10) + By – 38.5 = 0 By = + 60.5 kN By = 60.5 kN – 0.8(10) + Dx = 0 Dx = + 8 kN Dx = 8 kN Fx 0 : Part DC: Fx 0 : – 8 + Cx = 0 Cx = + 8 kN Cx = 8 kN 8 kN Part AE: Fx 0 : 8 – Ax = 0 Ax = +8 kN Ax = 8 kN + Fy 0 : 6 – Ay = 0 Ay = +6 kN Ay = 6 kN E 6 kN A Ax Ay 28.5 kN 34 kN.m 8 kN/m 4 kN/m B H E C D 8 kN 3m 8 kN A 4m 6 kN 4 60.5 kN 4 34 kN 4 3m Final Reactions - The force in the link member AE = 10 kN (Tension) With my best wishes Dr. M. Abdel-Kader 15 kN Question (2): (8 Marks) 3 4 8 kN/m a 9 kN b c 32 kN 28 kN 2m 4m 12 kN 8 kN/m (a) b a 9 kN 9 kN c 32 kN x = 2.5 m 28 kN 2m (b) Location and value of M max +ve 9 9 + Zero shear at: Fy left 0 32 – 12 – 8 x = 0 4m 0 N.F.D (kN) 0 x = 2.5 m 32 2 M max +ve = 32 (2.5) – 12 (0.5) – 8 (2.5) /2 Or 16 + = 49 kN.m 4 0 S.F.D (kN) 0 - Zero shear at: Fy right 0 2.5 m 8 x – 28 = 0 x = 3.5 m from support b M max +ve = 28 (3.5) – 8 (3.5)2/2 -28 0 B.M.D (kN.m) 0 = 49 kN.m + M max +ve = 49 kN.m at 2.5 m from the support a. 28 40 48 49 Mmax +ve Question (3): (3 Marks) F F Zero Parallel reactions (1) Stable + Determinate. (2) Unstable. (3) Unstable. With my best wishes Dr. M. Abdel-Kader
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