1 / 11

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

Mechanics of Materials – MAE 243 (Section 002) Spring 2008. Dr. Konstantinos A. Sierros. Problem 2.4-14

talbot
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 2.4-14 A rigid bar ABCD is pinned at point B and supported by springs at A and D (see figure). The springs at A and D have stiffnesses k1 = 10 kN/m and k2 = 25 kN/m, respectively, and the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm, respectively. A load P acts at point C. If the angle of rotation of the bar due to the action of the load P is limited to 3°, what is the maximum permissible load Pmax?

  3. Problem 2.5-5 A bar AB of length L is held between rigid supports and heated nonuniformly in such a manner that the temperature increase ΔT at distance x from end A is given by the expression ΔT = TB (x^3/L^3), where ΔTBis the increase in temperature at end B of the bar (see figure). Derive a formula for the compressive stress c in the bar. (Assume that the material has modulus of elasticity E and coefficient of thermal expansion .)

  4. 2.6: Stresses on inclined sections • Up to now we have considered normal stresses acting on cross-sections • Provided that the bar is prismatic, the material is homogeneous, the axial force P acts at the centroid of the cross-sectional area and the cross-section is away from any localized stress concentrations

  5. 2.6: Stress elements • Isolating a small element C (fig. 2-30 c) • We have a stress element (fog 2-31 a) • The only stresses acting are the normal stresses σx • Because it is more convenient, we usually draw a 2-D view of the stress element (fig. 2-31 b)

  6. 2.6: Stresses on inclined sections • In order to obtain a more complete picture, we need to investigate the stresses acting on inclined sections (plane pq fig. 2-32 a) • Uniform distribution of stresses as shown in figs 2-32 b and c

  7. 2.6: Stresses on inclined sections • We need to specify the orientation of the section pq. Define an angle θ between the x-axis and the normal n to the section • We need to find the stresses acting on the section pq • Load P, which is the stress resultant, can be resolved with respect to N and V • N is associated with normal stresses and V is associated with shear stresses

  8. 2.6: Stresses on inclined sections • Establish standard notation and sign convention • Normal stresses σθ are positive in tension and shear stresses τθ when they tend to produce counterclockwise rotation of the material where σx = P/A, in which σx is the normal stress on a cross-section

  9. 2.6: Stresses on inclined sections

  10. 2.6: Stresses on inclined sections • Element A: The only stresses are the maximum normal stresses (θ = 0) • Element B: This is a special case where all four faces have the same magnitude normal and shear stress (σx/2)

  11. 2.6: Stresses on inclined sections The shear stress may cause failure if the material is much weaker in shear than in tension

More Related