Method of sections Method of Sections It is
- Slides: 25
Method of sections
Method of Sections • It is based on the principle that if the truss is in equilibrium then any segment of the truss is also in equilibrium. • In this method, a section is passed through the truss to cut it into two parts so that internal forces are exposed at the cut members. • Equations of equilibrium are then applied to the free body diagram of either of the two parts to determine the desired forces. • In choosing a section of the truss, in general, not more than three members whose forces are unknown may be cut, since there are only three available equilibrium equations which are independent. • The method of sections has the basic advantage that the force in almost any desired member may be found directly. November 3, 2020 2
An Important Note If a body is in equilibrium under the action of external forces (including support reactions) then its each and every part is also in equilibrium under the action of external and internal forces. 11/3/2020 3
Problem -1 Compute the forces in the members DC, DB and AB of the truss, shown here, using method of sections. 11/3/2020 4
Support Reactions 11/3/2020 5
Calculation of the member forces For convenience, assume tension in all the cut members. 11/3/2020 6
Calculation of the member forces 11/3/2020 7
Member forces 11/3/2020 Member Forces (k. N) Method of Joint Forces (k. N) Method of section Nature AB 2. 0 Tension DC 2. 0 Tension DB -2. 5 Compression 8
Example 6: Method of Sections 6 -9
Example 6 (continued): 6 - 10
Example 7: Method of Sections 6 - 11
Example 7 (continued): 6 - 12
Problem For the loaded truss shown in the Figure : 1. Calculate the support reactions at A and F. 50 k. N 40 k. N 2 m 2 m 2. Identify the zero force members. 2 m D E 3. Determine the force in the member FE and FG using Method of Joints. 4. Calculate the force in the members ED, EH and GH using Method of Sections. Assume the supports at A and F as pin and roller respectively. November 3, 2020 F G H I C B 1. 5 m 30 k. N 1. 5 m 40 k. N 1. 5 m A 19
Solution of Part 1 (Support Reactions at A and F) 40 k. N 2 m 2 m 50 k. N 2 m D E F G C H Ax 30 k. N 1. 5 m B I Fy 1. 5 m A 40 k. N 1. 5 m Ay 11/3/2020 20
Solution of Part 2 (Zero Force Members) 40 k. N 2 m 2 m Only member GE is a zero force member as members FG and GH are collinear and there is no external force acting at joint G. 50 k. N 2 m D E G F C H Ax 30 k. N 1. 5 m B I Fy 1. 5 m A 40 k. N 1. 5 m Ay 11/3/2020 21
Solution of Part 3 (Forces in the members using Method of Joints) 40 k. N 2 m 2 m FFE F θ 2 m D E FFG Fy= 57. 5 k. N 50 k. N G F C H Ax 30 k. N 1. 5 m B I Fy 1. 5 m A 40 k. N 1. 5 m Ay 11/3/2020 22
Solution of Part 4 (Forces in the members using Method of Sections) 40 k. N 2 m 2 m E F G D H 40 k. N 2 m 2 m FED FGH 50 k. N 2 m D E 1. 5 m θ FEH G F Fy = 57. 5 k. N C H Ax 30 k. N 1. 5 m B I Fy 1. 5 m A 40 k. N 1. 5 m Ay 11/3/2020 23
- Method of sections
- Method of section
- Method of sections
- Advantages and disadvantages of symposium
- Geometric solids with circular cross sections
- Think on these things
- Declaration of independence sections
- Declaration of independence sections
- Ctd structure
- Parts of stomach
- Me-111
- Types of section views in engineering drawing
- Cutaways and cross sections definition
- Removed sectional view
- What is the voice box called
- Cobol 1959
- Pga sections map
- Parabolas in life
- Rotating conic sections
- Lesson 1 exploring conic sections
- Aristotle anatomy
- Semicircle cross section volume formula
- Application of parabola in real life
- Erg table 1
- Directional terms
- Cross section slice