Exam 2 Study Guide

What is a neutral axis?

Thursday, March 8, 2018

5:47 PM

Exam Info:

Will be on Tuesday, March 20th at 8-9:30pm in WIHR 200

• 3 questions long

· Any calculator accepted as long as it does not have internet connection

What are the shear and moment diagrams?

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

The axle of an automobile is acted upon by what?

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

Non-symetic 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 counter-clockwise 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) Ig-iż 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) .2267x10-3

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

1-9.69(0.1)

17.2)(-0.2) LM2710-3

-2.25

MPa

Ox

1.667x10

.2267x10-3

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/1O-03) 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 in-plane

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 = (1.5-1.632/6.018 = 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=ro-t = 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