SOLID STRUCTURES 1 & 2 PIN-JOINTED FRAMES(NEW 2023), Exams of Engineering Physics

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SOLID STRUCTURES 1 & 2 PIN-JOINTED FRAMES(NEW 2023)

SOILIDS AND STRUCTURES

2 - PIN-JOINTED FRAMES

1.155P (C) = 6kN → Pmax = 5.2kN Schedule 2022-23 Semester 1 (revised) Week Thursday 3-5pm (YOLC-LT1) 1 Intro to module 2 Equilibrium at a point 3 Equilibrium of a rigid body 4 Employability 5 Pin-jointed frames – method of joints

APPLICATION

Long trusses are often used to construct large cranes and electrical transmission towers. The method of joints requires that many joints be analyzed before we can determine the forces in the middle of a large truss.

So another method to determine those forces would be helpful.

THE METHOD OF SECTIONS

Today’s Objectives : You will be able to determine:

THE METHOD OF SECTIONS

In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude as the forces at

the joint. This result is based on the equilibrium principle and Newton’s third law.

STEPS FOR ANALYSIS

1. Decide how you need to “cut” the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general).

assume all the members are in tension, as done when using the method of joints (unless it’s obvious it has to be compression). Upon solving, if the answer is positive, the member is in tension as per the assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)

STEPS FOR ANALYSIS (continued)

5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!

ANSWER

H B

B. MB only involves 600 N applied force and

member force FHG. Lines of action of forces

FBG and FBC pass though B.

EXAMPLE

Given: Loads as shown on the truss and ignoring self-weight. Find: The force in members KJ, KD, and CD. Plan:

a) Determine the support reactions at A. b) Take a cut through members KJ, KD and CD. c) Work with the left part of the cut section. Can also work with right part of cut. Try it! d) Apply the E-of-E to find the forces in KJ, KD and CD.

a

Analyzing the entire truss for the reaction at A, we get F = 0: A = 0. A moment equation about G to find AY results in:

• MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 56.7 kN X X

EXAMPLE (continued)

56.7 kN

b/c

x (+) y (+) M (+)

Now take moments about point D. Why do this?

• MD = 56.7 (9) – 20 (6) – 30 (3) + FKJ ( 4 ) = 0 FKJ = − 75 kN or 75 kN (C)

d