Transcript Document

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 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 106 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?
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 δC of point C, located at
distance h from the lower end of the bar.
(b) What is the elongation δB 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?
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
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
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
FIG. 2-15
Statically indeterminate bar
Copyright 2005 by Nelson, a division of Thomson Canada Limited
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
(1)
• 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.
FIG. 2-16
Analysis of a statically
indeterminate bar
Copyright 2005 by Nelson, a division of Thomson Canada Limited
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)
FIG. 2-16
Analysis of a statically
indeterminate bar
Copyright 2005 by Nelson, a division of Thomson Canada Limited
2.4: Statically indeterminate structures – Procedure of analysis
• Next we need to use the equation δ = (PL)/(EA) in order to obtain the forcedisplacement relations for each segment
•Therefore we have:
(3)
(4)
…because segment CB
shortens…
FIG. 2-16
Analysis of a statically
indeterminate bar
Copyright 2005 by Nelson, a division of Thomson Canada Limited
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
δAB = δAC + δCB = 0
(2)
(3)
http://www.youtube.com/watch?v=addXg-iDSfc&feature=related
(4)
Monday 11 February 2008 (during class)
QUIZ on Statically indeterminate structures