< Final Exam Study Guide

What are the transformations of the plane stress?

Wednesday, May 2, 2018

4:29 PM

Exam Info:

Will be on Saturday, May 5th, at 1pm - 3pm in 4 questions long, the last of which is bonus points

Any calculator accepted as long as it does not have internet No notes/ book or any other materials allowed Will cover chapters 6.1,7.1-7.66,8.1, and 10.1 - 10.3

What are the stress transformations?

connection

notes

w

oments

12 on

STA

Equations

Given on exam

Moments of Ihertia

I recongle = 1 bh? I shaft = 2 C We also discuss several other topics like Who is immanuel kant and what are his philosophies?

Jeshaft = 2 2 2 Euler's famula

Pc: TEPE

• Stresses in a beam or a shaft

What is beam failure?

o me - C = The

wala We also discuss several other topics like Ethnology

. Stresses on

a pressure vessel

Ohoop

het

- Transformation of stresses

On - Oqt or rose cos 20 + Lid sin 28 Zxys 3. 004 sinze - Exy cos20

'. 0cten Ov-0 * cos 20 - Zoy sin 20 We also discuss several other topics like What is the meaning of the word symptom?

z

Emax = 10 ) + Exy Elastic Curve If you want to learn more check out Carbohydrates are made of what?

EI di = M(x)

LEP

Key Concepts

Chapter 61 - Horizontal Shearing Stress in Beams If you want to learn more check out What type of biomolecules are sex hormones?

If sher between

- sections

- glue keeps sections toogerher 2000

28000

- Iglue force F o r F, = { O, ddl - for yda'

M o MAM . S odd' = M A M JdA

니

UA

AX

(201$- Słuc

Fn = F2-F,

AM S. geht'

Fu a S yedd')

HULLE

"Formula of Ch.7" Oly) - So geht Ely) - Sydy vy? If you want to learn more check out What is ecliptic?

of

plane stress

•Chapter 7.1 - Transformation

- Coordinate Rotation

Principal normal stress

Maximum shear stress,

Tmax = 1 Omar

On: C. cose

On = Ox costo

N = Pioso

Me Renda

AA area of inclined surface

AA

Acoso

On = OxcoseO Oy Sinza

ty Nacost

Gayatrine

+ 2 y sin cost

Oglasima

{F = 7 Kent S -10-07) subcosé + Psylcore time)

trig shit 2 sin cos - Sin20 Sin? (1-10520) Cose B = {{1+cos20)

•Chapter 7.2 -

Mohr's

Circle

Stress Transformations

yield criteria

I max < of

Principal

sresses

Tresca's Hexagon

++

Mutu

3rd principal Stress

Emax = o-o con

O. - Ozzog O2=0,- og

y = x-xo

A

principal stresses

•Chapter 10 -

Beam

Failure

EM!:0 =xsin = K-20

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Ch.1-4

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h

like your knee

wwwww

y = A sindx

- Bersex

to do

StudySoup

BO

A sin th=0

[T = 0, = 0]

sinth to

Hoon

NT

:

L = n T

n: 1, 2, 3 ... -

[x = L y = 0) B

2= (n I2

NNNNN

P

(b)

A on the shaft is shown.

Example Questions

© Due to the applied loading the element at point

1) the in-plane principal stresses C 2) the in-plane maximum shear stress 3) drow the in-plane Mohr's circle 4) calculate the max out-of-plane shear stress r) draw Mohr's circle corresponding to ")

12 ksi

2 For the beam and loading shown with centroid C, derermine:

i) the shear force v at non. 2) the moment of inertia of the cross section 3) the shear stresses at B and B 4) the shear stress at C 5) sketch the shear stess distribution along the

vertical axis

80 IN

80 KN

StudySoup

ower

3 The W8+ 31 steel beam shown is to be used as a planed column.

For Oy = 50 ksi and E= 29 +10° psi determine 1) The rectangular moment of inertia appropriate

for buckling. 2) The largest centric load it can support before

the steel yields 3) the lorgest centric load it can support before

it buckles 4) the F.S. for a zoo ki load 5) What would the F.S. be if the column

was cantilevered?

12 ft So

Solutions

va =

اکی)

SOU

StuSOU

(1) aug, R

here

0.: -12ksi

Doug = - least

RE 1 (zobett? = 936436 = 8.4965

Tmax = 8.4 aasi

Og=0

uksi

Y

Studio

9.49

4) 7.25

2.5651

SENSO

2 Du=80kn

2) t = ve

Istal 15*2005) + 2( tz( soox 209 + 300 x 20x1105)) = 155.6x10“ Mm4 Q: (300x20)(110): 0.6602106 m3 4; (50.000w)(0.66On lommes), 22.6 MPa

(155.6x10..) (15mm) CB 1.13 MP : = 300mm

4)

0310.160x10.mes) * (15-100%(60) = 0.7352006.ms

(80.000~10.735 x 10 mm) 25.2 MPa Ze 7155.620 mm (15mm)

www

ANOS

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3 W8x31

1) Ig = 37.1 in4 , d=9.13 in2

2) Os P[A :7 P=0A = 50 ks) * 9.13 in? = 456 kips 3) P = (I VEI - (44/144)(29x106m)(57.1.) - 512 kips

yield betur buckles 4) FS. - 456/200 > 2.28 î î

Penzion (E)(I) = (Pep) lu) = (512/4) = 128

F.S. - 128/200 - 0.64

Study SOM

dy Sou

SESSOU