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

Mechanics of Materials – MAE 243 (Section 002) Spring 2008. Dr. Konstantinos A. Sierros. Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter d 1 = 0.75 in. over one-half of its length and diameter d 2 = 0.5 in. over the other half (see figure). The

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

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

  2. Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter d1 = 0.75 in. over one-half of its length and diameter d2 = 0.5 in. over the other half (see figure). The modulus of elasticity E =30 x106 psi. (a) How much will the bar elongate under a tensile load P = 5000 lb? (b) If the same volume of material is made into a bar of constant diameter d and length 8.0 ft, what will be the elongation under the same load P?

  3. Problem 2.3-10 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E, and weight W hangs vertically under its own weight (see figure). (a) Derive a formula for the downward displacement δCof point C, located at distance h from the lower end of the bar. (b) What is the elongation δBof the entire bar? (c) What is the ratio β of the elongation of the upper half of the bar to the elongation of the lower half of the bar?

  4. 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

  5. 2.4: Statically indeterminate structures • Reaction and internal forces of structures (such as springs, bars and cables) can be determined solely by FBD’s and equilibrium equations • Such structures are called statically determinate structures • When dealing with statically determinate structures we do not need to know the properties of the material

  6. 2.4: Statically indeterminate structures • If bar AB is fixed at both ends, there are two vertical reactions RA and RB but only one equation (ΣFvertical = 0) • Such structures are called statically indeterminate • In order to analyze these structures we need to use additional equations. Equations that contain the displacements of the structure

  7. 2.4: Statically indeterminate structures – Procedure of analysis • Rigid supports at both ends for prismatic bar AB which is loaded by an axial load P at an intermediate point C • Using ΣFvertical = 0 we have;RA – P + RB = 0 • But we need an additional equation in order to determine the two unknowns • We can use a second equation based on the fact that a bar with fixed ends does not change in length. (1)

  8. 2.4: Statically indeterminate structures – Procedure of analysis • If we separate the bar from its supports (fig. 2-16 b) we have a bar with both ends free loaded by RA,RB and P • These forces cause a change in length by δAB • When both ends are fixed: δAB = 0 (2) Equation of compatibility (change in length must be compatible with the conditions at the supports)

  9. 2.4: Statically indeterminate structures – Procedure of analysis • Next we need to use the equation δ = (PL)/(EA) in order to obtain the force-displacement relations for each segment • Therefore we have: (3) (4) …because segment CB shortens…

  10. 2.4: Statically indeterminate structures – Procedure of analysis • Therefore by solving equations (1), (2), (3) and (4) we can get RA and RB RA – P + RB = 0 (1) δAB = 0 (2) δAB = δAC + δCB = 0 (3) (4) http://www.youtube.com/watch?v=addXg-iDSfc&feature=related

  11. Monday 11 February 2008 (during class) QUIZ on Statically indeterminate structures

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