MEEM2003_Lecture_Notes_PatrickWong

MEEM2003_Lecture_Notes_PatrickWong - MEEM2003 Mechanics...

Info iconThis preview shows pages 1–16. Sign up to view the full content.

View Full Document Right Arrow Icon
MEEM2003 Mechanics Assessment Tasks/Activities: Exam 60% Course work 40% Labs (20%), Case Study (15%), Test (5%). Recommended book: Mechanics of Materials , Beer, Johnston and DeWolf, 4th edition in SI units, McGraw-Hill, 2006. Mechanics of Materials , J.M. Gere, 5th edition, Brooks/Cole, 2001.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Objectives: To provide with the means of analyzing and designing various machines and load-bearing structures. To introduce the fundamental concepts of stress and strain analysis . My part: Mechanics of Materials
Background image of page 2
Introduction: An engineering design process usually starts with a question, “What do you want?” (A design specification). Example: We want to build a platform or stage being strong enough to support the heaviest student in this class. Specification: able to support maximum mass 100 kg. Items to be thought of: Materials (metal?, wood?, plastic?), Shape, Dimensions…
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
A simple example: We want to know the induced forces in all members
Background image of page 4
Free body diagram of frame ABC 1. The body ABC 2. All external applied forces 3. Reactions B C 30 kN A A x A y C x C y 0.8 0.6
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Three equilibrium equations: Σ M=0 and take moment @ point C A x (0.6) – (30000) (0.8) = 0 A x = +40 kN (1.1) Σ F x =0; A x + C x = 0 C x = A x = 40 kN (1.2) Σ F y =0 A y + C y – 30000 = 0 A y + C y = +30 kN (1.3) B C 30 kN A A x A y C x C y 0.8 0.6
Background image of page 6
Free body diagram of member AB 1. The body AB 2. All external applied forces 3. Reactions B 30 kN A A x A y B x B y 0.8 Σ M=0 and take moment @ point B A y (0.8) = 0 A y = 0 (1.4) Put into (1.3) C y = 30 kN
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Free body diagram of frame ABC We found A x = 40 kN A y = 0 C x = 40 kN C y = 30 kN B C 30 kN A A x A y C x C y 0.8 0.6
Background image of page 8
Free body diagram for member BC B C C x C y 0.8 0.6 C x = 40 kN C y = 30 kN BC is a two-force member. Its own weight can be neglected. Reactions at the end points must be equal and opposite and collinear. F BC = (C x 2 + C y 2 ) = 50 kN
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
B C F BC F’ BC C F BC F’ BC B D F BC F’ BC D D The internal force at different sections on member BC is always F BC (= 50 kN tension).
Background image of page 10
B C 30 kN A A x C x C y 0.8 0.6 FBD of point B B 30 kN F AB F BC FBD of member AB’ B’ A F AB = A x F’ AB Internal force in member AB is F AB ( = A x = 40 kN compression)
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
If the cross sectional area of boom AB is A AB = 30 x 50 mm 2 The induced normal stress in boom AB is σ AB = normal force ÷ cross sectional area = 40x10 3 /(30x50x10 -6 ) = 26.7 MN/m 2
Background image of page 12
Similarly, the diameter of bar BC is 20 mm and it area is then A BC = π . 20 2 / 4 = 314.2 mm 2 The induced normal stress in bar BC is σ BC = normal force ÷ cross sectional area = 50x10 3 /(314.2x10 -6 ) = 159.1 MN/m 2 120 Brass 216 Carbon Steels Yield Stress (MPa) Material Which material would be the right choice?
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Stress and Strain Analysis Direct stress Stress at section XX A F = = area sectional Cross load Normal σ This assumes that the force is uniformly across the cross section. SECT XX F Tensile F Cross-section area= A F Compression F
Background image of page 14
If the force is not uniformly distributed over the cross section, Small element of area, dA, carrying load, d F .
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 16
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/10/2009 for the course MXXM 2XX9 taught by Professor Gxxy during the Spring '09 term at City University of Hong Kong.