Understanding Load, Shear, and Bending Moments in Engineering Mechanics
Explore the relationships between load, shear, and bending moments in statics through examples and problem-solving techniques. Learn how to plot shear and moment diagrams for simple beams, calculate reactions, and analyze desired sections with detailed solutions. Dive into SF and BM calculations to gain a comprehensive understanding of structural mechanics.
Download Presentation
Please find below an Image/Link to download the presentation.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author. If you encounter any issues during the download, it is possible that the publisher has removed the file from their server.
You are allowed to download the files provided on this website for personal or commercial use, subject to the condition that they are used lawfully. All files are the property of their respective owners.
The content on the website is provided AS IS for your information and personal use only. It may not be sold, licensed, or shared on other websites without obtaining consent from the author.
E N D
Presentation Transcript
Relations Among Load, Shear and Bending Moment Reactions at supports, wL = = R R A B 2 Shear curve, x 0 = = = V V w dx wx A wL 2 L 2 = = = V V wx wx w x A Moment curve, + x 0 = = M M Vdx _ A x ( ) L w 0 = = 2 wL M w x dx L x x 2 2 + + 2 dM = = = at 0 M M V max 8 dx 5- 3
Problem-1 Plot the shear and moment diagrams produced in the simple beam by the forces shown. 4 kN 2 kN/m 2.8 kN.m B A 1 m 1 m 1 m 1 m 1 m 5/2/2025 5
Solution Reactions: 4 kN 2 kN/m 2.8 kN.m A x A B 1 m 1 m 1 m 1 m 1 m y A B y + = = 0 0 F A x x + = + = + = 0 2 1 4 0 6 F A B A B y y y B y y + = ) 1 5 . 2 + = ( ) 0 5 4 4 2 ( 8 . 2 0 CCW M A y = = = . 3 64 kN 6 . 3 64 . 2 36 kN B A y y 5/2/2025 6
Desired sections 4 kN 2 kN/m 2.8 kN.m x 0 . 1 0 A x A B x 1 0 . 2 1 m 1 m 1 m 1 m 1 m . 3 64 kN x 2 0 . 3 . 2 36 kN x x x 0 . 3 0 . 4 x x x 0 . 4 0 . 5 x 5/2/2025 7
SF and BM calculations x 2.8 kN.m 0 . 2 0 . 3 2 kN/m x 0 . 1 x 0 0 . 1 0 . 2 V V 2.8 kN.m V A A M o o A M x M o x 2 m x . 2 36 kN . 2 36 kN . 2 36 + kN + = + = 0 0 Fy Fy = 0 Fy = = . 2 V 36 = 0 . 2 V 36 = ( 2 ) 2 0 V x V = . 2 V 36 = 0 V (Constant) kN 36 . 2 . 6 36 2 (Linear x in ) x (Constant) kN 36 . 2 + = ( ) 0 + CCW M + + = ( ) 0 CCW M + = O ( ) 0 CCW M O O . 2 36 8 . 2 x M x = . 2 36 0 M x + = . 2 36 8 . 2 0 M x ( ) 2 = . 2 36 (Linear x in ) M x = = 8 . 2 + ( 2 ) 2 0 x . 2 36 (Linear x in ) M x 2 = + 2 6 36 . 6 8 . M x x (parabolic in ) x 5/2/2025 8
SF and BM calculations x 0 . 4 0 . 5 x 0 . 3 0 . 4 2 kN/m 4 kN 2 kN/m 2.8 kN.m M 2.8 kN.m V V M A A o o 1 m 1 m 1 m 1 m 1 m 1 m 1 m x x . 2 36 Fy kN = . 2 36 kN + = 0 Fy + 0 = . 2 V 36 = 2 1 C 0 V = . 2 V 36 = 2 1 4 0 V . 0 36 ( onstant) . 3 (Constant) kN 64 + = ( ) 0 CCW M + = ( ) 0 CCW M O O + + ( 1 = . 2 36 8 . 2 + 2 ) 5 . 2 x 0 M x x + + ( 1 ) 5 . 2 + = . 2 36 8 . 2 + 2 ( 4 ) 4 0 M x x x = . 0 36 2 . 2 (linear in ) M x = . 3 64 18 2 . (linear in ) M x x 5/2/2025 9
SF and BM calculations (contd.) = V 0 . 1 M = M (Constant) kN 36 . 2 . 2 36 (Linear x + in ) x x 0 = 8 . 2 . 2 36 (Linear x in ) x = (Constant) kN 36 . 2 V x 1 0 . 2 = + 2 . 6 36 6 8 . (parabolic in ) M x x x = . 6 36 2 (Linear x in ) V x x 2 0 . 3 = 2 . 2 + . 0 36 (linear in ) M x x V = . 0 36 ( onstant) C x 0 . 3 0 . 4 = . 3 + 64 18 2 . (linear in ) M x x = (Constant) kN 64 . 3 V x 0 . 4 0 . 5 x 0 . 1 x x x x 0 0 . 1 0 . 2 0 . 2 0 . 3 0 . 3 0 . 4 0 . 4 0 . 5 x 0 .5 1.0 1.0 1.5 2.0 2.0 2.5 3.0 3.0 3.5 4.0 4.0 4.5 5.0 2.36 2.36 2.36 2.36 2.36 2.36 2.36 1.36 0.36 0.36 0.36 0.36 -3.64 -3.64 -3.64 SF kN 0 1.18 2.36 -0.44 0.74 1.92 1.92 2.85 3.28 3.28 3.46 3.64 3.64 1.82 0 BM kNm 5/2/2025 10
Shear force and Bending Moment Diagrams 4 kN 2 kN/m 2.8 kN.m A B 1 m 1 m 1 m 1 m 1 m . 3 64 kN . 2 36 kN . 2 36 . 0 36 SFD . 3 64 . 3 64 . 3 28 . 2 36 . 1 92 BMD . 0 44 5/2/2025 11
Sample Problem 2 SOLUTION: Taking entire beam as a free-body, calculate reactions at B and D. Find equivalent internal force-couple systems for free-bodies formed by cutting beam on the left and right sides of load application points. Draw shear force and bending moment diagrams for the beam and loading shown. Plot the results. 5- 12
Sample Problem 2 - Continued SOLUTION: Taking free-body of the entire beam, calculate the reactions at B and D (By=46 kN and Dy=14 kN). Find equivalent internal force-couple systems at sections on both sides of load application points. 0 kN 20 V = : y F 1= = 0 20 kN V 1 ( )( ) M 1= +M 1= 1= 0 : 20 kN 0 m 0 0 M Similarly, = = 20 kN 50 kN m V M 2 2 = + = 26 kN 50 kN m V M 3 3 = + = + 26 kN 28 kN m V M 4 4 = = + 14 kN 28 kN m V M 5 5 = = 14 kN 0 kN m V M 6 6 5- 13
Sample Problem 2 - Continued Plot results. Note that for a beam subjected to concentrated loads, shear force is of constant value between concentrated loads and bending moment varies linearly. + _ _ Remember Sign Convention V + M M + _ _ V 5- 14
ENGINEERING MECHANICS : STATICS Sample Problem 3 5- 15
ENGINEERING MECHANICS : STATICS Sample Problem 3 - Continued + 0 m < x < 6 m 6 m < x < 10 m _ + + 5- 16
ENGINEERING MECHANICS : STATICS Sample Problem 4 5- 17
ENGINEERING MECHANICS : STATICS Sample Problem 4 - Continued 5- 18
ENGINEERING MECHANICS : STATICS Sample Problem 4 - Continued 5- 19
ENGINEERING MECHANICS : STATICS Sample Problem 4 - Continued 0.233 1.799 At x=2 m , M=1.799 kN.m At x=2.23 m, M=1.827 kN.m (maximum moment) 5- 20
ENGINEERING MECHANICS : STATICS Sample Exam Question 5- 21
ENGINEERING MECHANICS : STATICS Sample Exam Question 5- 22
Sample Problem 6 SOLUTION: The change in shear between A and B is equal to the negative of area under load curve between points. The linear load curve results in a parabolic shear curve. With zero load, change in shear between B and C is zero. The change in moment between A and B is equal to area under shear curve between points. The parabolic shear curve results in a cubic moment curve. Sketch the shear force and bending-moment diagrams for the cantilever beam and loading shown. The change in moment between B and C is equal to area under shear curve between points. The constant shear curve results in a linear moment curve. 5- 23
