SOLID STRUCTURES 2 PIN-JOINTED FRAMES, Exams of Engineering Physics

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SOILIDS AND STRUCTURES

2 - PIN-JOINTED FRAMES

3. If the maximum force that any member of the truss in Fig. Q3 can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D. [P = 5.2kN] Approach:
   1. Whole structure equilibrium to determine reaction forces Ay and Ey
   2. Analyse forces at pin D
   3. Analyse forces at pin E (taking advantage of symmetry)
   4. Analyse forces at pin C
   5. Compare max tensile and compressive member forces with criteria Tmax = 8kN and Cmax = 6kN to determine Pmax

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 (^6) Pin-jointed frames – method of sections (^7) STUDY WEEK (^8) Friction (^9) Stress

(^10) Stress (^11) Axial deformation (^12) Compatibility INTERIM ONLINE ASSESSMENT (20%)

APPLICATION

Forces in truss members using the method of sections.

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.

3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.

STEPS FOR ANALYSIS (continued)

4. Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially, you may 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

B. MB only involves 600 N applied force and

member force FHG. Lines of action of forces

FBG and FBC pass though B.

H 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.

Analyzing the entire truss for the reaction at A, we get FX = 0: AX = 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

EXAMPLE (continued)

56.7 kN

b/c

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

a

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)

EXAMPLE (continued)

Now use the x- and y-direction equations of equilibrium.

d

56.7 kN x (+) y (+) M (+)

Video worked solution on Canvas Determine the force in members BC, CF, and FE. Take P 1 = 480 kN and P 2 = 780 kN.

Supporting reading/reference

Clifford – Intro to ME:

Part 1 Unit 1 Solid

Mechanics

Hibbeler – Statics &

Mechanics of

Materials

Blockley – Structural

Engineering: A Very

Short Introduction

• Ch.5 pp.257- 262 - Self-study:
• Try self-study problems (at end of this PPT)
• Watch video – pin-jointed frame2 (on Canvas)