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LECTURE 31

Dr. Mustafa Y. Al-Mandil Department of Civil Engineering. LECTURE 31. 1. Internal Forces ( Beams & Frames. 71. Chapter 7. 50 N. 4. 3 m. 3. A. B. C. 5 m. 5 m. 50 N. V. M. M. N. N. V. R Cx = 30 N. R Ay = 20 N. R Cy = 20 N. Dr. Mustafa Y. Al-Mandil Department of

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LECTURE 31

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  1. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 31 1 Internal Forces ( Beams & Frames 71 Chapter 7 50N 4 3m 3 A B C 5m 5m 50N V M M N N V RCx = 30N RAy = 20N RCy = 20N

  2. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering INTERNAL FORCES: These internal Forces depend on:- 1: Geometry of member. 2: Loading of member. 3: Location within member. A M N=0 V 2m 20N LECTURE 31 2 72 Chapter 7 ( N ) 1- Axial (N) or (A) 2- Shear (V) 3- Moment (M) ( V ) ( M ) 40N 2m e.g. A B 6m

  3. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering 20N/m B A 2m 5m 20N/m M A N 2m V 50N LECTURE 31 3 73 EXAMPLE 1 Chapter 7 (1): Find external Reactions. (4): Find Internal Forces. 20N/m (2): Make Section at required location. 0 0 50N 50N (3): Select smaller part of sectioned member.

  4. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Shear & B.M. Diagrams: M N x V R Chapter 7 1 - Draw F. B. D. for Beam. 2 - Solve for External Reaction. 3 - Determine Number of Segments. 4 - Decide on Global Coordinate System. 5 - Make Section @ x of each Segment. 6 - Solve for Internal Forces as Function of ( x ): N = f ( x ) V = g ( x ) M = h ( x ) 7 - Draw Internal Forces Vs. (x). 8 - Determine Max. & Min. Values of Functions 9 - Good Luck ! V M N x R

  5. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering + Mx - 60x == 0 Chapter 7 20N/m Using F · B · D : B A B = 0 Ray = 60N () 6m Fy = 0 + RBY = 60N () F · B · D 120N 20N/m RBx A B M Fx = 0  RBx = 0 RBy N RAy A Take Section @ x: N 60 V Fx = 0  N = 0 Fy = 0 +  60 - 20 x - V = 0 x + VN 60N _ 90 -60  V = 60 - 20x + MN·m  M = 60 - 10x2

  6. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering + + 20N/m x Chapter 7 EXAMPLE: 20N/m A Using F·B·D. B 6m RBy 60N F·B·D. RBx M 2m x N MB A NN V +60 Parabola + VN MN _ Hyperbola -120N·m

  7. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering 50N Example - 3 4 3 M N 40N 30N A x M B x A N 20N V V 20N Chapter 7 A C Using FBD for ABC: Fx = 0  Rcx = 30N ( c = 0 RAy = 20N ( Fy = 0  Rcy = 20N ( SEGMENT AB ( 0 x < 3 ) B 40N 30N B C Rcx A 3m 3m RAy Rcy Fx = 0  N = 0 Fy = 0 V = 20N Mx = 0  M = 20x SEGMENT BC ( 3 < x 6 ) x _ NN -30 -30 Fx = 0  N = 0 - 30N (c) Fy = 0 V = - 20N Mx = 0  M = 20x - 40 (x - 3) = 120 + 20 x - 40 x = 120 - 20 x +20 +20 + _ -20 -20 +60 + MN·m

  8. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Example M A N B V x 15N M N V Chapter 7 20N 5N/m 80N·m C B A MA = 0  RBY = 35N Fy = 0 RAy = 15N 30N 20N 80N·m SEGMENT AB ( 0 <x < 6 ) 35N 4m 6m 15N V = 15 - 5x M = 15x - +15 +20 +20 + + _ V(N) SEGMENT BC ( 6 <x < 10 ) -15 x 5N/m 22.5N·m +80N·m 35N 15N x - 6 + 6m + MN·m V = 20N M = 20x - 120

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