Introduction
The most general state of stress at a point may
be represented by 6 components,
x , y , z
normal stresses
xy , yz , zx shearing stresses
(Note : xy = yx , yz = zy , zx = xz )
Same state of stress is represented by a
different set of compo
Homework #7
Due Tuesday, November 1, 2011
Shear Stresses in Beams Pages 374-377
7-4, 7-7, 7-8, 7-9
7.4 ans. max = 499 psi, jump = 332 psi
7-8 ans. Vweb = 27.1 kN
7-7 and 7-8 are considered 1 problem for grading
Shear Flow in Built-up Members Pages 384-386
Homework #5
Solution to be Posted Tues., Sept. 20, 2011
This assignment will not be collected for a grade.
This material will be covered on Exam 1 scheduled for
September 27.
Calculation of torsional stress pgs. 193-199
5-1, 5-15, 5-19
Calculation of angl
Homework #6b
Due Tuesday, October 18, 2011
The Flexure Formula pages 294-301
6-53, 6-54, 6-55, 6-77, and 6-85
6-54 and 6-55 will be treated as one problem and may therefore
utilize the same sketch, given statement, and find statement.
Composite Beams page
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Homework #9
Solution to be posted Tues., Nov. 29, 2011
Stress Transformation Equations Pages 453-460
9-16, 9-30, 9-38, 9-39
The answer to 9-16 is the same as 9-54
Mohrs Circle Pages 424-431
9-54, 9-73, 9-78
1- 1
Homework #3
Due Tuesday September 6, 2011
Relationships and Hookes Law pgs. 98-101
3-10, 3-11, 3-15
and Relationships pgs. 111-112
3-29, 3-33
Generalized Hookes Law
See the problem on the next page
(Ans. a'=2.00302", b'=2.00553", t'=0.24964)
1- 1
Genera
Stability of Structures
In the design of columns, cross-sectional area is
selected such that
- allowable stress is not exceeded
=
P
all
A
- deformation falls within specifications
=
PL
spec
AE
After these design calculations, may discover
that the col
Deformation of a Beam Under
Transverse Loading
Relationship between bending moment and
curvature for pure bending remains valid for
general transverse loadings.
1
=
M ( x)
EI
Cantilever beam subjected to concentrated
load at the free end,
1
=
Px
EI
Cur
Introduction
Transverse loading applied to a beam
results in normal and shearing stresses in
transverse sections.
Distribution of normal and shearing
stresses satisfies
Fx = x dA = 0
Fy = xy dA = V
Fz = xz dA = 0
(
)
M x = y xz z xy dA = 0
M y = z x dA
Centroids of Areas
Q y = x A = xA
Qx = y A = yA
Symmetry
Centroid lies on axis of symmetry
2 axes of symmetry
Point symmetry
1 axis of symmetry
Centroids of Common Shapes
Inside front cover of textbook
Composite Areas
Qy = X A = xA
Qx = Y A = y A
Example
Mechanics of Solids CE 220
University of Alabama at Birmingham Fall, 2011
l. A rod passes through a hole in the ceiling and is
supported by a disk afxed to its end as shown.
Determine:
a. The required diameter (d) of the rod if the rod has
an ultimat
Homework #6a
Due Tuesday October 11, 2011
Centroids and Moments of Inertia
1. Determine y which locates the centroid, then find
the moments of inertia Ix' and Iy for the T-beam.
Ans. - y = 206.82 mm, Ix' = 221.6(106) mm4, Iy = 115.1(106)mm4
2. Determine t
Homework #1
Due Tuesday August 30, 2011
Calculation of Stresses Pages 39-45
1-34, 1-38, 1-51, 1-65
For 1-65 also determine the bearing stress under the triangular
blocks
Design of Simple Members Pages 55-59
1-75, 1-85, 1-89, 1-100
1-100 Answers t = 5.33 m
Homework #12
Solution to be posted by Thurs., Dec. 1, 2011
Beam Deflections by Integration Pages 587-592
12-10, 12-27
Beam Deflections by Superposition Pages 624-626
12-92, 12-102
Ans. 12-92 is A = 0.0190 rad, C = 25.3mm
Indeterminate Beams by Integration