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 x 10 6 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 δ C of point C, located at distance h from the lower end of the bar. (b) What is the elongation δ B of 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 members are 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 R A and R B but only one equation (ΣF vertical = 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 ΣF vertical = 0 we have; R A – P + R B = 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 R A, R B and P These forces cause a change in length by δ AB When both ends are fixed: δ AB = 0 Equation of compatibility (change in length must be compatible with the conditions at the supports) (2)

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: …because segment CB shortens… (3)(4)

10. 2.4: Statically indeterminate structures – Procedure of analysis Therefore by solving equations (1), (2), (3) and (4) we can get R A and R B R A – P + R B = 0 δ AB = 0 δ AB = δ AC + δ CB = 0 (1) (2) http://www.youtube.com/watch?v=addXg-iDSfc&feature=related (3)(4)

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