Mechanics of Solids I

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Jul 18, 2012 (5 years and 3 months ago)

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ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
OK, Midterm stuff:
Class average was huge! = 86% -including 26 aces (100s)!
Much better results than last year
Beam Diagrams were very well done.
A few had a bad day –don’t worry you can make it up in the final.
Now lets go over it . . .
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Problem 1
10
marks
Correct answer 5 marks
Correct formulas 5 marks
The two cylindrical rods AB and BC are welded together at B and loaded as shown.
Knowing that d1 = 50 mm and d2 = 30mm, find the average normal stress at the
midsection of rod AB and rod BC
40 kN
30 kN
BONUS for 5 marks: What will be the total extension of the two cylinders if AB is brass
with E = 101 GPaand BC is Aluminum E = 68.9 GPa?
At d1 Normal stress = force/area
Sigma = 70000/(3.14*.025^2) = 35.7 MPa
At d2 Normal stress = force/area
Sigma = 30000/(3.14*.015^2) = 42.4 MPa
Delta = sum(PL/ AE) = (70000*.3)/(3.14*.025^2*101*10^9) +
(30000*.25)/(3.14*.015^2*68.9*10^9)
= 0.000262m
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Problem 2
20
marks
Correct answer 10 marks
Correct formulas 10 marks
The torques shown are exerted on pulleys A, B, and C.
Knowing that both shafts are solid determine the maximum
shearing stressin shaft AB and shaft BC
400 Nm
1200 Nm
800 Nm
Ans= 75.5 MPaand 63.7 MPa
Shear stress=Tau= Tc/J
Torque in AB = 400 Nm
J of AB = ½pi*c^4
J = 0.5*3.14*.015^4= 7.95*10^-8
C of AB = 0.015
Therefore
Tau(AB) = 400*.015/7.95*10^-8
= 75 MPa
Torque in BC = 800 Nm
J of AB = ½pi*c^4
J = 0.5*3.14*.02^4= 2.5*10^-7
C of AB = 0.02
Therefore
Tau(AB) = 800*.02/2.5*10^-8
= 64 MPa
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Problem 3
20
marks
Correct answers 10 marks
Correct formulas 10 marks
The torques shown are exerted on pulleys A, B, and C. Knowing that
both shafts are solid and made of brass (G=39 GPa), determine the
angle of twistbetween A and B, and, A and C.
400 Nm
1200 Nm
800 Nm
(same sketch as Problem 2)
Twist (rad) = TL/(JG)
Twist(AB) = 400x1.2/(7.95x10^-8x39x10^9)
Twist(AB) = 0.155 radians or 8.88 deg -ccw
Twist(BC) = 800x1.8/(2.5x10^-7x39x10^9)
Twist(BC) = 0.147 radians -cw
Twist(AC) = 0.155-0.147 = .007 radians or 0.40 deg
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Problem 4
30
marks
0.5m
L = 2m
w = (M
+N
+C
)/L
Draw shear and bending moment diagrams and determine the maximummoment for each load case (see data sheet for loading)
0.5m
0.5m
M
N
C
Correct sketches 5 marks each = 20
Maximum moment 5 marks each = 10
L = 2m
CASE 1
CASE 2
(WL^2)/8 = 20.625
41.25c
-41.25
38.0
V
M
26.8
-20.0
-44.5
32.4
V
M
R1 = R2 = (11.22+46.86+24.42) / 2 = 82.5/2 = 41.25 kN
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
20
marks
Problem 5
If the beam in Problem 4 was an 8 inch x 5 inch aluminum beam with the properties indicated on the data sheet, determine the
absolute maximum bending stress in the beam for both load cases.
If the beam were oriented on its sideas shown below instead of the upright position, what would be the absolute maximum
bending stress for both load cases.
Side orientation:
Correct answers 5 marks each = 20
BONUS for 5 marks: Will the beam fail in either case, why or whynot?
Sigma(xx) = M(max)*Cxx/ Ixx
20.6*101*10^6/28x10^6 = 73.6 MPaor 10.6 Ksi
32.4*101*10^6/28x10^6 = 115.7 MPaor 16.7 Ksi
Is Sigma greater than Sigma(ultimate) = 45 KSI = 311 MPa?
no, no for I yes, yes for H
Sigma(yy) = M(max)*Cyy/ Iyy
20.6*64*10^6/3.6x10^6 = 370 MPaor 53 Ksi
32.4*64*10^6/3.6x10^6 = 582 MPaor 84.4 Ksi
Ixx= 28x10^6 mm^4
Iyy= 3.6 x10^6 mm^4
64mm
101 mm
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
For the aluminum beam in question 4 and 5, what would be the maximum P(ultimate) load it could support before failing if the
load was applied at a point in the middle of the simply supported beam as sketched below?
HERE’S THE DEAL!!
I have purchased this aluminum beam
and it is in the structures lab. We are
going to break it in the lab next week!
The student that gets closest to the
actual ultimate bending load in this test
will be awarded a cash prize equal to the
value of the scrap aluminum (probably
around $50), or, will be given a bonus of
20% on their midterm mark (not to
exceed 100%) –their choice.
P(ultimate)
BIG BONUS PRIZE
Answer = I don’t know. It’ll be interesting to see.
I’m going to check bending, direct shear and transverse shear
Bending Max: M=PL/4 gives us M=500P when L=2000 mm
Sigma= M y / I and Sigma ultimate is 310 MPawhile Ixx= 28x10^6mm^4 and y = 101mm
Therefore: 310 = 500*P*101/28x10^6 gives us P of 172 kNfor bending
Check direct shear:
area = 3853 mm^2 shear ultimate = 30 KSI = 207 MPa= 799 kNshear resistance –no probshere!
What about transverse shear?
tau= VQ/It = 207=V*[(96.4*127*10.4) + (45.6*91.2*6.4)]/(28*10^6*6.4) therefore, V = 241 kNand thus P = 482 kN
So it will fail in flexure and I’d say that the imperfections and local flaws will be cancelled by the conservative thicknesses
from manufacturing therefore stick with the 172kN = just over twice the weight of the entire class!
127 mm
10.4 mm
91.2 mm
6.4 mm
Potential = 20 extra marks
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
LOADS
Approximate engr-4312 Student Enrolment:
Navals= 17, Mechanicals = 71, Civils= 37
TOTAL of 125 students
Assuming the average student weight = 0.66 kN
Then:
N
= 17*0.66 = 11.22 kN
M
= 71*0.66 = 46.86 kN
C
= 37*0.66 = 24.42 kN
8”ALUMINUM I -BEAM PROPERTIES
8”ALUMINUM I -BEAM PROPERTIES
CONVERSION FACTORS
Inch = 25.4 mm
KSI = 6.9 MPa
DATA SHEET for BEAM QUESTIONS
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Chapter 7: Transverse Shear
In this chapter we will develop a method
of finding the shear stress in a beam.
Also, shear flow, will be discussed and
examples will be worked.
Shear in a beam subject to bending may
be longitudinal and transverse.
Longitudinal can be illustrated by the
bending beam below:
If the boards are bonded then shear stresses build
up and the cross section warps. This condition
violates our assumption of sections remaining plane
when bent but warping is relatively small especially
for a slender beam.
We will now use the assumptions or homogeneity
and prismatic cross section to develop a shear
formula similar to the flexure formula. . .
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Shear Formula
It is important to recall that shear stress is
complimentarymeaning transverse and
longitudinal shear stresses are numerically
equal
.
The derivation and proof of the shear formula
are detailed in HibbelerChapter 7 section 2.
Review this in your own time because I want to
get straight to the formula and how it is used.
It
VQ
=
τ
The shear stress in the member at the point located y’from the neutral axis.
This stress is assumed to be constant and therefore averaged across the
width t of the member.
The Internal resultant shear force, determined from the method of sections
and the equations of equilibrium
The moment of inertia of the entire cross sectional area computed about the
neutral axis.
The width of the members cross sectional area, measured at the point
where is to be determined
τ
V
I
t
'''
'
AydAyQ
A
==

τ
Where A’is the top or bottom portion of the member’s
cross sectional area, defined from the section where t
is measured, and is the distance to the centroidof
A’, measured from the neutral axis
'y
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Shear Formula
It is necessary that the material behave in a linear elastic manner and have a
modulus of elasticity that is the same in tension as it is in compression.
It
VQ
=
τ
Shear Stresses in Beams
Applying the shear formula for common beam cross-sectional situations:
Rectangular:
()
[]
()
[]
bbh
byhV
It
VQ
3
22
12/1
4/
2
1







==
τ








−=
2
2
3
4
6
y
h
bh
V
τ
A
V
5.1
max
=
τ
A
V
avg
=
τ
Maximum shear acts
on the neutral axis
(centerline here) and
near the ends where
V is greatest.
Note -Parabolic
Parabolic
Transverse Shear
Hibbeler
Chapter 7
Note that b is now
removed (check
earlier notes)
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Rectangular Beam continued:
So that it can be well understood it can be shown that integrating the shear
stress, , over the entire cross-sectional area A yields the shear force V.
τ
坩摥⁆污湧攠䉥慭猺
䄠睩摥⁦污湧攠扥慭⁣潮獩獴猠潦⁴睯⁦污湧敳⁡湤⁡⁷敢⸠䅮⁡湡汹獩猠潦⁴桥⁳桥慲=
楮⁡⁷楤攠晬慮来⁢敡洠牥獵汴猠楮⁴桥⁩汬畳瑲a瑩潮⁢敬潷t
Parabolic
Jump due to the smaller “t”in
the shear formula
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
In summary the important points to remember about shear stressesin
beams are:
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Examples:
Transverse Shear
Hibbeler
Chapter 7
It
VQ
=
τ
'''
'
AydAyQ
A
==

Key Formulas for Solution:
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
'''
'
AydAyQ
A
==

ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
'''
'
AydAyQ
A
==

ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
It
VQ
=
τ
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
It
VQ
=
τ
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Examples:
Transverse Shear
Hibbeler
Chapter 7
It
VQ
=
τ
'''
'
AydAyQ
A
==

ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
Shear Flow in Built up Members
Built-up members are often used in engineering
applications. If loading on built-up members causes
bending then fasteners are usually required to keep
the pieces from sliding over each other. Nails,
screws, glue, bolts, welds etc must then resist the
shear at along the length of the member. This shear
loading along the member is called SHEAR FLOW
and is computed as a force per unit length.
q = the SHEAR FLOW (force per unit length along the beam)
V = Shear Force at that section of beam
I = moment of inertia of ENTIRE cross section
Q =
'''
'
AydAyQ
A
==

Where A’is the top or bottom portion of the member’s
cross sectional area, defined from the section where t
is measured, and ybaris the distance to the centroidof
A’, measured from the neutral axis
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
Shear Flow in Built up Members
I couldn’t say it any better than Hibbelerso read the following:
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
Examples of Shear Flow
433
231.23)75.4)(5.5(
12
1
)25.5)(6(
12
1
inI=−=
3
75.3)25.0)(6(5.2''inAyQ===
The beam will fail at the glue
joint for board b since Q is a
maximum for this board.
)25.0)(2(231.23
)75.3(
400;
V
It
VQB
allow
==
τ
kiplbV24.11239
=
=
N
A
5.5 in
0.25 in
0.25 in
0.25 in
5 in
2.5 in
2.5 in
a
b
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Transverse Shear
Hibbeler
Chapter 7
Examples of Shear Flow
433
231.23)75.4)(5.5(
12
1
)25.5)(6(
12
1
inI=−=
3
75.3)25.0)(6(5.2''inAyQ
b
===
psi
It
VQ
B
b
646
)25.0)(2(231.23
)75.3)(10(2
3
===
τ
N
A
5.5 in
0.25 in
0.25 in
0.25 in
5 in
2.5 in
2.5 in
a
b
3
4375.3)25.0)(5.5(5.2inQ
a
==
psi
It
VQ
a
a
592
)25.0)(2(231.23
)4375.3)(10(2
3
===
τ
0.25 in
ENGI 4312
Mechanics of Solids I Prof. Steve Bruneau, EN.4013
Ph 737-2119
sbruneau@engr.mun.ca
T.A.s Rizk, Dawood, Guo
rizk@engr.mun.ca
nabil@engr.mun.ca
chaoguo@engr.mun.ca
COURSE
TEXT READING
INSTRUCTOR
TOPIC
NOTES
Combined Loadings & Thin-Walled vessels
This chapter serves as a review of the stress-analysis that has been
developed in the previous chapters regarding axial load, torsion,
bending and shear. The solution to problems where several of these
loads occur simultaneously will be studied. Prior to this, the stresses in
thin-walled vessels will be analyzed.
Combined Loadings
& Thin Walled Vessels
Hibbeler
Chapter 8