ME311 Strength of Materials (Winter, 2012)
Problem Set #4 Solution
Problem 1
Problem 2
3 2
1
i Ai xi yi
i Ix
1 10.06 0.5 0.03 (1.0)(0.06) 3 12 (0.06)(1.0) 3 12 0.5-0.274 0.03-0.401
2 (1-0.12)0.06 0.03 0.5 (0.06)(0.88) 3 12 (0.88)(0.06) 3 12 0.03-0.274 0.5
ME 311 Winter 2013
Problem Set #11
Due Friday, April 12, 2013
Problem 1
The rigid bar AB is attached to a hinge at A and to two springs, each of constant k. Use
the stability criterion based on potential energy to determine the relation between m, k d,
h
Practice problems
1.
The Z-section below has second moments of area Ix = 560, 000 mm4, Iy = 290, 000
mm4, Ixy = 300, 000 mm4. It is used for a beam of length 2m which is simply supported at
its ends and loaded by a uniformly distributed vertical load of 1
ME311 Strength of Materials
(Winter, 2012)
Problem Set #2
(Due: Friday, January 20)
Problem 1
A uniform beam of weight w per unit length is supported by concentrated forces at A and
B. The cross-section is T- shaped, having the centroid and relevant dimen
ME311 Strength of Materials
(Winter, 2012)
Problem Set #3
(Due: Friday, January 27)
Problem 1
Determine the stress at point A for the beam with the cross-section shown below. Show
the orientation of the zero stress line.
Ix = 185.9 in4, Iy = 1730 in4, Ixy
1
3.30. Use Castiglianos first theorem to solve Problem 3.18.
From the geometry of the system, we have a sin = b sin and hence
b cos d = a cos d or
d
a cos
=
.
d
b cos
The spring is relaxed when = 30o , so the spring compression is
= 2a sin 2a sin 30o
ME 311 Winter 2013
Problem Set #3 Due: Friday, January February 1
Problem 1
The beam shown below has a spring support at end A. The force - deflection relation for
the spring is F = k " where k is the spring constant and ! is its length change.
Find (i) T
ME 311 Winter, 2013
Problem Set #5 Due: Friday, February 15
Problem 1
Consider a cantilever beam having the cross-section shown below, and having a force P
acing at its free end. The area properties are:
I y = 0.0219 in4 , Iz = 0.0118 in 4 , Iyz = 0.00672
ME 311 Winter, 2013 Problem Set #7 Due: Friday, March 1 Problems 1 and 2 In the structures shown below, the bars are connected by smooth pins. The bars have the same area A and elastic modulus E. Their lengths can be determined from the figure. (a) State
ME311 STRENGTH OF MATERIALS
Fall 2016
COURSE OUTLINE
Date
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
Class
W 9/7
F 9/9
M 9/12
W 9/14
F 9/16
M 9/19
W 9/21
F 9/23
M 9/26
W 9/28
F 9/30
M 10/3
W 10/5
F 10/7
M 10/
ME 311 STRENGTH OF MATERIALS
Practice Examination I
Attempt all questions.
1. (65 points) Figure 1(a) shows a thin circular beam of flexural rigidity EI and radius A
which extends for 360o (2 radians). The beam is built in (clamped) at C and a force F is
1
3.43. Figure P3.43 shows a stepped steel shaft supported in bearings at A,C and
loaded by a 2 kN force at B. Use Castiglianos second theorem to find the angular
misalignment at the bearings (i.e. the local slope of the beam) due to the shaft
deflection.
1
12.6. A uniform simply-supported beam of flexural rigidity EI and length L is
subjected to a lateral uniformly distributed load w0 per unit length and an axial
force P as shown in Figure P12.6. Find the maximum bending moment if P is
equal to 70% of the
1. No calculations are required for this question. I mean it! No calculations.
Please write the answers to this question on this page.
(i) Sketch the approximate location of the principal axes for the beam section shown in
Figure 1. Label the stiff and ex
Practice problems Solution
1.
The maximum bending moment will occur at the mid-point and can be determined from
the free-body diagram. We find
Maximum tensile stress will occur at the point C defined by the
coordinates.
Substituting these values into the
Practice Problems
1.
Two springs join at a point. Find the displacement component u F of the joint point due to
the inclined force F. Show that it reaches a minimum when = 105. (Note: spring
constant k is similar to S)
uF
2. Use appropriate energy method
Practice Problems Solution
Problem 1
Problem 2
Problem 3
F(3)
D 1 A P1
F (4) = P
F(4)
B Q
P
F (3) = -Q
2
3 B P
F(1) F
(2)
A F(3)
P1
- F (3) - F (2) / 2 = 0 P - F (1) - F (2) / 2 = 0 1 F (2) = 2Q F (1) = P - Q 1
C
4 Q
Bar 1 2 3 4
1/S() L AE 2L AE L AE L AE