1 / 8

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

mollie-duran
Download Presentation

LECTURE 31

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related