Chapter 7 Internal Forces ( Beams & Frames LECTURE 31 1 71 50N 50N 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

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

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

Chapter 7 Shear & B.M. Diagrams: x x 1 - Draw F. B. D. for Beam. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Chapter 7 Shear & B.M. Diagrams: 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 V R x M N x R

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

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

Chapter 7 Example - 3 50N A 20N _ _ Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Chapter 7 4 3 50N Example - 3 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 A Rcx A V N M x 20N 3m 3m RAy Rcy Fx = 0  N = 0 Fy = 0 V = 20N Mx = 0  M = 20x SEGMENT BC ( 3 < x 6 ) x _ NN A x V N M 20N B 40N 30N -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

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