Description
Exam 2 Study Guide
Thursday, March 8, 2018
5:47 PM
Exam Info:
Will be on Tuesday, March 20th at 89:30pm in WIHR 200
• 3 questions long
· Any calculator accepted as long as it does not have internet connection
Don't forget about the age old question of What is uncertainty avoidance?
• Noc notes book or any other materials allowed
• Will cover chapters"4,5, and a We also discuss several other topics like What is qc?
Equations :
Given on exami
Bending: Oss Mante We also discuss several other topics like What is the embryonic stage?
I= bh2/12
Moment of Inatia
(vestangle) If you want to learn more check out What are exogenous waves?
ale
Parallel axis theorem: Je!  In + Ad
Key Concepts
Centroids of beams:
I Area  y
 Ayr L
T
Miten
Eyedi & Ai
y
Study BOU
Nonsymetic sending
Studs
y compress
compression
Onze rico
max
tension
neatcal exis beams bend in
one direction more than others
So
tand = tan 0 1
Shear & moment diagrams: (le rules)
O Concentrated forces causes a jump (discontinuisy) in shear Don't forget about the age old question of What is ancestral homology?
diagram at the location of the force. same direction as the force arrow
irta
I
3
The change in shear forces between any two locations is equal to the area under the distributed lood dlagrom The slope of the shear diagram at any location is equal to the distributed load value Imagnitude and sign) at that location
4 The change in moment between any two locations We also discuss several other topics like What did the wagner act do?
is equal to the area under the shear diagram 3 The slope of the moment diagram at any location
is equal to the shear force at that location
©
Concentrated external moments cause a jump (discontinanity) in the internal bending moment at the point of application.
A clockwise external moment causes diagram to jump up. A counterclockwise external moment causes diag. to jump down.
· Deformations in Beams:
Emax = Ex: +ş :
(normal strain is @ max when y=c) Emas
Base
9   
curatat
leasiest

problem
there
is
 Ig
has
solving to
Сос
 PlL)
CEI til 2 =PL  px EI de la =PLx  Porno
EI 46)  PL per extremo
EI O'r)  PLx  pet
boundry conditions:
4:0 a 20 ivo EI y(t) = Pc 
· Indeterminant Beams (method of super position) :
use
table
given
cantileo
LILIT
unting
ang
simple
simply supported
dont combine
cantilever
and
simply
supported
cases
study sour
Special Casesi
(Extended
Beams)
no bending
Study
Sud SOU
Vc = sino
no bending at B to
in BC, so you just find the displacement
Stud
need at C
the
slope
Lecture 26
Friday, March 9, 2018
7:35 AM
First
problem 
Ch.u
unsu metic
beadines
5) Igiż bh?: [1.4)(.29) = 0.2664 4103
aj Iz izbh. Izl.2.4) = 1.664 v10°34
c)
Me = }(2): 7.2kw.mm My = $(12) = 96 kv.m
M:12knom
(72)(0.2) 1.067x103
(4.0) fo.1) .2267x103
2. IMPO
14.05 MB
5x10
14.95 MPa
1000 kPa
ec: 0.12100), cena to), 4.4540*ve a) CD: 0.10cm), cocco) + 12.25 Air
19.69(0.1)
17.2)(0.2) LM27103
2.25
MPa
Ox
1.667x10
.2267x103
W = 5 kn/m VILIT
3
5u
5
A4C40
~ A: 5.75 kw
<< < 0  0  (5)  25(5) • lo c =  34 35 x
연
3425
Sas t
25
34.25 4
Sadd beat
It
쑖
(M
2 са
3 unknowns
8 kip
case it trick
case 2
Wo Zxipl3+
case I (flipped)
55
Ye + Osl
Yo = y. + OBL
de a
167
69
IGET
24EI
Marletta) { = could : 40 xipfe
A+B=28 EMA=0 ?
A=18 kips B: 10 kips
(5:18)+8
40
o = M. 740
7. 6
3
Given: Diagram as shown
2 si
Find: (a), le principal planes
(btle principal stresses
Solution
(a)
20,= 24y/1O03) 0 = 3 ton (zzgl.16.0)
02  18.435 Ou  18.435° +90° l. = 108.4359)
(b) Omex = or more 1/20) + tay
3:270* 1/23  11 Omin Ox toy (oros Congo
Ower = lles!
= 2100 + 2/(27077 (3)2 =
min = 1 ksi)
7.10
 Given : Diagram as shown
CKS
ksi
Find: (a) the orientation of the planes of maximum inplane
shearing Stress (b) the maximum shearing Stress
(the corresponding normal stress Solution
(a) osa į tau a oriental  Ź tou / 2017 26.6
10. = 26.6°) 9.  Os +90o =26.6° +909 2=63.4°
MA
(b) Comes to geng)  Yay : 19")*(9)"=5
Tmax = 5 ksi)
Dong = lessi)
7.23
o
Given: The axle of an automobile is acted upon by the
forces and couple shown. Diameter of axel = 32 mm
SO
0.15m
0.2m
150 Nem
3k
WW
Find: (a) the principal planes and principal stresses at painer H
(6) maximum shearing stress at point H
Solution CE 32.mk2 = Ilomma Olom
J = 40"/2= 6.06 )4/2 = 1,0289410 *m* I: YC4/4 = 26.06.14/2 = 5.1445 x 108 m4 Exy = Tc/5 = (30)21.016)/(1.0289~09)  54.426 MPa M = .15mF = 0.15 * 3000w = 450 Nm 4 eggs out . 134.65 AP.
42O
U
lf 39,955
10 = 18.90
o
e = @provo
139.955
lomax 18.67 MPa Omin or to o.  t t = 7 Lomin=  158. COMPA)
Care = Jo  Trang => [{mos = 88.6 MPa)
44
LAMOK
7.106
SO
Given A bulk Storage tank has an outer diameter of
3.3m and a wall thickness of 18 mm. Internal
pressure of the tank is 1.5 MPa Find : maximum normal stress and re maylium sheer stress
Solution
pedo t= {(3.3)  0.018 = 1.632m
G= prit = (1.51.632/6.018 = 36 MPa
10: 136 MPa]
mar: pr/2 : (1.5.1.6327(2.0.018) = 68 MPa
[Tmax = 68MPa]
7.117
Given: The Pressue tank below hos a 0.375 in wall thickness
and butt welded seams forming an angle 3.20° with a transverse plone. Guage pressure = 85 poi
a
transwer
sure
r
15ft
Find: (a) normal stress perpendicular to the weld . (b) the shearing stress parallel to the weld
Solution:
de = 54 60m Todo/2 = 30:n r=rot = 30.375 = 29,625 O = Pele = (85•29.625)/0.375 = 6715psi On = Pofle aiko, : {(1715): 3357.5 psi © Taug = 10.702)/2 = 5036.25psi Yomon : lo.  de)/2 = 1678.75psi
Ow = Oang  Cmax cos(2p)  5036.25  1678.75c05(40) = 3750.25 psi
Ow= 3750.3 psi
(b)
: Emak sinl2)) = 1678.5 sin (40)  1079.07 psi
= 1079.07pi
tudy Sour