1 / 12

Mechanics of Materials – MAE 243 (Section 002) Spring 2008

Mechanics of Materials – MAE 243 (Section 002) Spring 2008. Dr. Konstantinos A. Sierros. Problem 1.8-6

kiet
Download Presentation

Mechanics of Materials – MAE 243 (Section 002) Spring 2008

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Dr. Konstantinos A. Sierros

  2. Problem 1.8-6 A suspender on a suspension bridge consists of a cable that passes over the main cable (see figure) and supports the bridge deck, which is far below. The suspender is held in position by a metal tie that is prevented from sliding downward by clamps around the suspender cable. Let P represent the load in each part of the suspender cable, and let θrepresent the angle of the suspender cable just above the tie. Finally, let σallow represent the allowable tensile stress in the metal tie. (a) Obtain a formula for the minimum required cross-sectional area of the tie. (b) Calculate the minimum area if P 130 kN, 75°, and allow 80 MPa.

  3. Problem 1.8-9 A pressurized circular cylinder has a sealed cover plate fastened with steel bolts (see figure). The pressure p of the gas in the cylinder is 290 psi, the inside diameter D of the cylinder is 10.0 in., and the diameter dbof the bolts is 0.50 in. If the allowable tensile stress in the bolts is 10,000 psi, find the number n of bolts needed to fasten the cover.

  4. Homework Set 1 • 4) Answer true or false and justify each answer in one sentence. • You can measure stress directly with an instrument the way you measure temperature with a thermometer FALSE. Stress is an internal quantity that can only be inferred but cannot be measured directly (b) If a normal stress component puts the left surface of an imaginary cut in tension, then the right surface will be in compression FALSE. Tension implies that the normal stress pulls the imaginary surface outward, which will result in opposite directions for the stresses on the two surfaces produced by the imaginary cut (c) The correct way of reporting positive axial stress is σ = +15 MPa FALSE. The normal stress should be reported as tensile

  5. Homework Set 1 • 4) Answer true or false and justify each answer in one sentence. • (d) 1 GPa equals 106 Pa • FALSE. 1 GPa equals 109Pa • (e) 1 psi is approximately equal to 7 Pa • FALSE. 1 psi nearly equals 7 kPa and not 7 Pa (f) A common failure stress value for metals is 10000 Pa • FALSE. Failure stress values are in millions of pascals for metals • (g) Stress on a surface is the same as pressure on a surface as both quantities have the same units • FALSE. Pressure on a surface is always normal to the surface. • Stress on a surface can be normal or tangential to it

  6. 2.1 Introduction Chapter 2: Axially Loaded Members • Axially loaded membersare structural components subjected only to tension or compression • Sections 2.2 and 2.3 deal with the determination of changes in lengths caused by loads • Section 2.4 is dealing with statically indeterminate structures • Section 2.5 introduces the effects of temperature on the length of a bar • Section 2.6 deals with stresses on inclined sections • Section 2.7: Strain energy • Section 2.8: Impact loading • Section 2.9: Fatigue, 2.10: Stress concentration • Sections 2.11 & 2.12: Non-linear behaviour

  7. 2.2 Changes in lengths of axially loaded members - Springs • Spring stretching or shortening is analogous to the behaviour of a bar under axial loading • If the material of the spring is linearly elastic, the load (P) and elongation (δ) are proportional • P = k δ and δ = f P • Therefore; • k = 1/f and f = 1/k stiffness flexibility http://www.youtube.com/watch?v=xqANn3sYv18

  8. 2.2 Changes in lengths of axially loaded members –Prismatic bars • A prismatic bar is a structural member having a straight longitudinal axis and a constant cross-sectional area throughout its length • However, we can have a variety of cross-sections • If we assume that material is homogeneous and linearly elastic then elongation δ is: • δ = (PL)/(EA) axial rigidity Stiffness = (EA)/L and flexibility = f = (L)/(EA)

  9. 2.2 Changes in lengths of axially loaded members – Cables • Cables are used transmit large tensile forces (eg. lifting heavy objects, supporting suspension bridges etc) • They cannot resist compression and have little resistance to bending • Cross sectional area of a cable = total cross sectional • area of individual wires • Total cross sectional area of individual wires is called effective area or metallic area

  10. 2.2 Changes in lengths of axially loaded members – Cables • When under the same tension, the elongation of a cable is greater than that of a solid bar since the wires in the cable ‘tighten up’ • The effective modulus of the wire rope is less than the modulus of the material it is made • In practice the properties of the cables are obtained from the manufacturers • For solving problems in this book, use Table 2.1

  11. Please study example 1.8 (pages 46-48) • Next time we will discuss about it. • Therefore, you need to make sure that you have comments/questions ready for discussion

  12. Monday 4 February 2008 during class… Quiz covering Chapter 1 Duration: 20 mins Solve 1 out of 2 Questions

More Related