Statics Chapter 7
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The document summarizes key concepts from a statics lecture, including internal forces (axial, shear, moment), how they depend on member geometry and loading, and the process for analyzing beams and frames to determine internal forces as functions of location. It provides an example of using the force-balance diagram (FBD) method to analyze a beam with varying loads and calculate the shear, moment and axial force at any point as equations in terms of location x.
Statics Chapter 7
- 1. CE 201 - STATICS LECTURE 31 1 71 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Internal Forces ( Beams & Frames 50N 50N 3m 5m 5m A B C 3 4 M M N N V V RAy = 20N RCy = 20N RCx = 30N
- 2. CE 201 - STATICS LECTURE 31 2 72 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering 1- Axial (N) or (A) 2- Shear (V) 3- Moment (M) ( N ) ( V ) ( M ) These internal Forces depend on:- 1: Geometry of member. 2: Loading of member. 3: Location within member. mN N y x 40M0ΣM 20V0ΣF 0N0ΣF ⋅ =⇒= =⇒= =⇒= 2m 6m A B 40N 2m A N=0 M V 20N
- 3. CE 201 - STATICS LECTURE 31 3 73 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering 5m A B 2m 20N/m 0 0 20N/m 50N 50N A 2m 20N/m 50N M N V (1): Find external Reactions. (2): Make Section at required location. (3): Select smaller part of sectioned member. (4): Find Internal Forces. mN C N y x 6040100M 02501220M0ΣM 104050V 0V220500ΣF 0N0ΣF ⋅ =−=∴ =×−××+⇒= ↓=−=∴ =−×−⇒= =⇒=
- 4. CE 201 - STATICSDr. Mustafa Y. Al-Mandil Department of Civil Engineering 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 ! M N V R x M N VR x
- 5. CE 201 - STATICSDr. Mustafa Y. Al-Mandil Department of Civil Engineering Using F · B · D : + Ray = 60N (↑) ΣFy = 0 +↑ RBY = 60N (↑) F · B · D ΣFx = 0 ⇒ RBx = 0 ΣΜB = 0 Take Section @ x: ΣFx = 0 ⇒ N = 0 ΣFy = 0 + ↑ 60 - 20 x - V = 0 ∴ V = 60 - 20x ∴ M = 60 - 10x2 ΣMx - 60x == 0 2 x M N V 60N x 20N/m A 6m A B 20N/m 120N A B RAy RBy RBx N + _ 60 -60 VN MN·m 90 +
- 6. CE 201 - STATICSDr. Mustafa Y. Al-Mandil Department of Civil Engineering EXAMPLE: Using F·B·D. mN B B B N Byy Bxx 120M 0260M 0ΣM 60R0ΣF 0R0ΣF ⋅ =∴ =×− = ↑=⇒= =⇒= 6 20x mN3 x 2 y x 36 20x M 0 3 x 2 x 6 20x M 0ΣM N 6 10x 2 x 6 20x V0ΣF 0N0ΣF ⋅ − =∴ = −− = = ⇒= =⇒= + + M N V x A 20N/m 6m A B 20N/m 60N NN + _ +60 VN MN F·B·D. 2m x RBy RBx MB Parabola Hyperbola -120N·m
- 7. CE 201 - STATICSDr. Mustafa Y. Al-Mandil Department of Civil Engineering Using FBD for ABC: ΣFx = 0 ⇒ Rcx = 30N (←) ΣΜc = 0 ⇒ RAy = 20N (↑) ΣFy = 0 ⇒ Rcy = 20N (↑) ⇒ SEGMENT AB ( 0 ≤ x < 3 ) ΣFx = 0 ⇒ N = 0 ΣFy = 0 ⇒ V = 20N ΣMx = 0 ⇒ M = 20x ⇒ SEGMENT BC ( 3 < x ≤ 6 ) 4 3 50N A B C BA C Rcx 40N 30N 3m 3m RcyRAy x _ NN -30-30 +20+20 -20 -20 _ + + +60 MN·m A V N Mx 20N ΣFx = 0 ⇒ N = 0 - 30N (c) ΣFy = 0 ⇒ V = - 20N ΣMx = 0 ⇒ M = 20x - 40 (x - 3) = 120 + 20 x - 40 x = 120 - 20 x A x V N M 20N B 40N 30N
- 8. CE 201 - STATICSDr. Mustafa Y. Al-Mandil Department of Civil Engineering ΣMA = 0 ⇒ RBY = 35N ΣFy = 0 ⇒ RAy = 15N SEGMENT AB ( 0 < x < 6 ) SEGMENT BC ( 6 < x < 10 ) V = 15 - 5x M = 15x - 2 5 2 x A x V N M 15N B V N M x 15N 35N 5N/m x - 6 6m V = 20N M = 20x - 120 20N A B C V(N) -15 +20 +20 _ + + MN·m 5N/m 80N·m 30N 6m 4m 15N 80N·m 20N 35N + + 22.5N·m +15 +80N·m