SE110B Assignment 1
Due 04/11/2014
1. Wire AB of length L1 = 30 in. and wire BC of length L2 = 36 in. are attached to a
ring at B. Upon loading, point B moves vertically down by an amount
B = 0.25 in. Determine the extensional strains in wires (1) and (2

SE110B. Lecture 10 (Monday, 04/21/2014)
Torsion of noncircular prismatic bars
For a rectangular prism bar, the maximum shear stress is
T
d t 2
TL
=
, J = d t 3
GJ
max =
For an elliptical shaft
max =
=
2T
a b 2
TL
a 3 b 3
, J= 2
GJ
a + b2
Inelastic torsi

SE110B. Lecture 8 (Wednesday, 04/16/2014)
Thin-wall torsion members
Shear flow
V4 = A t A x = q A x , V2 = B t B x = qB x
V4 = V2 q A = qB
Therefore, q = t = const
dFs = qds
dT = dFs = qds
T = q ds
Cm
Cm
ds = 2 Am
T
2 Am
T
=
2tAm
Angle of twist
TL
ds
= 2

SE110B. Lecture 11 (Wednesday, 04/23/2014)
Chapter 5. Bending of Beams
_ _
P * Q * PQ
= lim x * x
x = lim
_
QP
x 0 x
PQ
_
_
A * B * = x = * , P * Q * = x* = ( y ) *
( y ) * *
y
=
*
Ey
x = lim
x 0
x =
Euler-Bernoulli beam theory.
x =
Ey
F

SE110B. Lecture 13 (Monday, 04/28/2014)
Unsymmetric bending:
Case 1. A beam with transverse loading parallel to an axis of symmetry.
Case 2. Doubly symmetric beams with inclined loads
x =
M yz
Iy
Mzy
Iz
Orientation of the neutral axis
My
I
y
* Mz *
z

SE110B. Lecture 15 (Monday, 05/05/2014)
Inelastic bending of beams
Elastic-plastic bending
Yield moment
I
M Y = Y , where c is the larger of c1 and c 2 .
c
MP =
Y A
2
(d C + dT )

SE110B. Lecture 20 (Friday, 05/16/2014)
Buckling load and buckling mode (for pined-pined conditions):
2 EI
Pcr = 2 ,
L
x
v( x) = C sin
L
Critical (buckling) stress
2 EI 2 E Ar 2
2E
=
=
,
cr Pcr / A =
AL2
AL2
(L / r )2
( )
r = I / A = the radius of

SE110B. Lecture 21 (Monday, 05/19/2014)
Effective length of columns
For elastic buckling load of columns with arbitrary end conditions:
2 EI
Pcr = 2
Le
Le KL
Eccentric loading:
EIv ' ( x) + Pv ( x ) = Pe
2
Let = P / EI
v ' '+ 2 v = 2 e
L
v( x ) = e tan

SE110B. Lecture 3 (Friday, 04/04/2014)
Shear stress and shear strain
Hookes law and Poissons ratio
x = E x , when 0 Y
y = z = x
Hookes law for plane stress; the relationship between E and G
1
( x y )
E
1
y = ( y x )
E
1
xy = xy
G
E
G=
2(1 + )
x =

SE110B Assignment 5
Due 05/9/2014
1. Aluminum-alloy cover sheets are bonded to a plastic core to form the sandwich
beam whose cross section is shown in the following figure. Determine the
maximum normal stresses in the aluminum skin and the plastic core.

SE110B Assignment 4
Due 05/2/2014
1. The shaft shown in the following figure is made of linearly elastic, perfect plastic
material. (a) Find its TY and TP . (b) Determine the radius rY of the elastic core
when T = 4 KN.m. (c) Determine the angle of twist

SE110B Assignment 8
Due 06/06/2014
1. A uniform steel rod with a diameter of d = 30 mm and a length L = 1.5 m is
subjected to an axial load P. Let E = 200 GPa and Y = 250 MPa. (a) Determine
the minimum value of P at which yielding occur. (b) Calculate the

SE110B Assignment 2
Due 04/18/2014
1. A steel pipe ( E = 200 GPa) surrounds a solid aluminum-alloy rod ( E = 70 GPa),
and together they are subjected to a compressive force of 200 kN acting on rigid
end caps. Determine the shortening of this bimetallic co

SE110B Assignment 7
Due 05/30/2014
1. As shown in the following figure, the force in spring BD is proportional to its
elongation, with spring constant k. At = 0 the spring is unstretched. Determine
the critical load, Pcr .
2. A steel pipe column is slende

SE110B Assignment 6
Due 05/16/2014
1. In the following cross section: (a) Locate the centroid and determine the moments
of inertia I y ' , I z ' , and the product of inertia I y 'z ' . (b) Determine the orientation of
the neutral axis.
2. An unequal-leg a

SE110B Assignment 3
Due 04/25/2014
1. The aluminum-alloy shaft AC (G=20 GPa) has an 800-mm-long solid segment
AB and a 400-mm-long tubular segment BC. The shaft is subjected to the
torsional loading shown in the following figure. The diameters are:
d 1 =

SE110B. Lecture 2 (Wednesday, 04/02/2014)
2.4 Elasticity and plasticity
Linearly elastic, perfectly plastic model
2.5 Linear elasticity; Hookes law and Poissons ratio
x = E x
y = z = x

SE110B. Lecture 4 (Monday, 04/07/2014)
Chapter 3. Axial deformation
Basic theory of axial deformation
Strain-displacement equation
u ( x + x) u ( x) du
x * x
lim
= x ( x) = lim
= x 0
= dx
x 0
x
x
Total elongation
e = u ( L ) u ( 0)
L
e = ( x)dx