SOILIDS AND STRUCTURES 2- PIN-JOINTED FRAMES, Exams of Engineering Physics

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Method of Sections

Dr Schleyer

Pin-jointed Frames

Solids and Structures 1 (ENGG110)

Or

‘Why things don’t fall down’

‘WE DO STRESS ANALYSIS TO DESIGN STRUCTURES SAFELY’

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:
4. Whole structure equilibrium to determine reaction forces Ay and Ey
5. Analyse forces at pin D
6. Analyse forces at pin E (taking advantage of symmetry)
7. Analyse forces at pin C
8. Compare max tensile and compressive member forces with criteria Tmax = 8kN and Cmax = 6kN to determine Pmax

1.155P (C) 1.155P (C) 1.155P (T) 1.155P (T) 1.155P (T) 0.557P (C) 0.557P (C)

P 2P

NOTE SYMMETRY

FBD of pin C 1.155P y FBC C x 1.155P

→ +  FX : (1.155P cos60) × 2 - FBC = 0 FBC = 1.155P kN (T) Criteria: 1.155P (T) = 8kN → P = 6.9kN 1.155P (C) = 6kN → Pmax = 5.2kN

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: Forces in truss members using the method of sections.

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).
2. Decide which side of the cut truss will be easier to work with (minimize the number of external reactions).
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.)

A. 3 unknowns. No need to

determine the support reactions.

ANSWER

H B

B. M

B

only involves 600 N applied force

and member force F

HG

. Lines of action of

forces F

BG

and F

BC

pass though B.

ANSWER

H B

FOR YOU TO DO

A) 90kN B) 57kN C) 45kN D) 23kN Given: Loads as shown on the truss and ignoring self-weight. Calculate the vertical support reaction at A?

EXAMPLE (continued)

56.7 kN 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)

a

b/c

d

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

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

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 -