ME 323: Mechanics of Materials

Fall 2020

Homework 11

Due Friday, December 04

Problem 11.1 (10 Points)

The elbow shown below is fixed to the ground at the center of the coordinate system. Two loads 75 Nand 150 N are applied at the free end in the y and z directions, respectively. If the elbow has a circularcross-section with a diameter of 20 cm, find:

(a) The internal reactions at a cross-section perpendicular to the y-axis at point A (y = 1 m). Classifythe forces as either axial or shear forces, and the moments as either bending or torsion.

(b) The stresses induced (magnitude and direction) in the stress elements M and N on the cross-sectionat A, shown in Fig. 11.1 (b), due to each reaction calculated in part (a). Use the three-dimensionalstress elements in Fig. 11.1 (c).

(c) Draw the Mohr’s circle for the stress states at M, and N, and calculate the principal stresses and theabsolute maximum shear stress |⌧max|.

1

Solutionsa

FBDYang

y

Fyi B.IEyNfyqoN

nm

Xz

9 Em

ETA 25 87 H

from

FIfxEtFyJtfIMAMxEtMyJ MzEFrom b

ZF 0

FI 1758 15017 0

as EA 75F 150 CN

INT O

MI RTAX 758 15055 0

NTA 28 8 755 15015oi.ME 1508 30051150 NM

Reaction at A Type ofreactionI 758N

15013NAxial force

2 shear force3 1508NM Bending moment4 300JNM Torsion5 150 NM Bending moment

bFor the circular cross sectionA 0212 _0.0314mL

I e.lt 4 7 854XIO

Sm4JIp TC0E5 t571XloYm4

1 758N axial force

8mci 8na 8y ff f.to 4 238p8a5

2 150 E N Shear force in E

Emp Ty z 434 4415073100314

6369.42PaIN 1 0

3 1508 Nm Bendingmoment

8N 2 82 1MXIRt 4383gl s

ln9lX1o5pq8MR8y O

4 300JNm Torsion

IMR ETE f 1 3,5949774

I 91 18paTNR tyx IMZRI 3 EEoy i9lxi8pa5 150 Nm Bending moment

8Nc3 By O

8mc3 8y IMEIk 7834977

I 91 X105Pa

In summaryThe state of stress at M

8M y 8mi I 18M 12T 8m13193 388 kpa

TM eYZ TimeI TMR 197.369 kpa193388kpa

y

369Rpg

ITt

The state of stress at N8N IT 2Nd 18N 2T 8N 3 193.388 Kp

INCDX T NI ITT Ne2 191kpa

ply19338819Pa

tea

ITt

Table method

ReactionatA Stress comp stress compcom ANH

I fy 75N 8yf2388j538y fAA238865A

2 Fz 150N Tyz zZ

636942Pa

3 Mx 150AM 8y Mx ft9lB4 My5300mm aoEI.IEplpaasxMERft9pHaSMz150AM8y fhZ f

L914105Pa

C form

Sarge 1932388 96694 kpa

F 9474497.369727219781Pa Max

8aVgyR 31647hPa

858mg RI 123.086KeaImafmin

82 0y

get3.381 19736

Eary gs8 kpa

Z veckpa

for N 8avg 9G 6941Pa12 214.08 KPa ITMax

81 310.775 8maX83 117.386 pa Grin 82 0

i

farads8CKPa

81ears 19L

HECKPD

ME 323: Mechanics of Materials

Fall 2020

Homework 11

Due Friday, December 04

Problem 11.2 (10 Points)

For the given stress element, determine the following:

(a) The principal stresses, the absolute maximum shear stress |⌧max| and the von-Mises stress.(b) The factor of safety using the maximum shear stress theory and the maximum distortion energy

theory for a ductile material with yield strength 32 ksi.(c) Whether or not failure is predicted using Mohr’s failure criterion for a brittle material with ultimate

compressive strength 90 ksi and ultimate tensile strength 30 ksi.

2

Solution se

Ex 20 KsiSy 10 Ks i

Tty8 Ksi

8avg8281 5 Ks i

R fT FTxj ffF A Rsi

Principal stressG Savoy R 12 Ksi

8mg R 22K si

Absolutemaximum shear stressImaxl R 17 Rsi

von Mises stress

8M FEET 29.866 Ksi

b Sy 3 2 Ksi

For maximum shear stress theorySF 8421TmaxT

O 94

For maximum distortion energytheorySf 848M 71.071

Mohris criterion805 90 Ksi807 30 less

R2g 0,824080C IT got z fourth quadrant

88k 704

82 0.24480C

Here EETFailure not predicted

ME 323: Mechanics of Materials

Fall 2020

Homework 11

Due Friday, December 04

Problem 11.3 (10 Points)

A horizontal rigid beam ACB is pinned to a wall at A and is supported by a pin-fixed column CD, as shownin the figure below. The beam is subjected to a linearly distributed load between B and C. The columnCD has Young’s modulus E and a rectangular cross-section, as shown. Consider supports C and D to actas pinned-fixed when buckling in the x-y plane and fixed-fixed when buckling in the y-z plane. Determinethe maximum linearly distributed load q such that column CD does not buckle. For this problem, use theparameters: L = 1.5m, b = 30mm, h = 45mm, E = 150GPa. Use Euler’s buckling theory to solve theproblem.

3

Solution

qfADFBD at

yc A FAX13k

3 FcKat

MAIO94 Fck O Fc 7zqLFor buckling inky plane

Iz fzhb3

Zr T

EfI11445011109 42 0045 0039

7 2 151233989 KN

andz 29L qcr 3P3 7

9cr 97HKN M

for buckling in y Z plane

Ix Iz bh3

Zr qEfITix 0111091 003140049

05 2 15 2

149.894 kN

Por 7 9L D 9 er 3Perp3 742 826 KNIM

so the maximum distributedloadof 9711 KNIM

ME 323: Mechanics of Materials

Fall 2020

Homework 11

Due Friday, December 04

Problem 11.4 (5 points)

1. Consider the state of stress shown below in a ductile material. Let ⌧MSS and ⌧MDE be the values ofthe shear stress ⌧ above that correspond to failure of the material using the maximum shear stressand maximum distortional energy theories, respectively. Circle the answer below that best describesthe relative sizes of ⌧MSS and ⌧MDE. (2.5 points)

(a) ⌧MSS < ⌧MDE

(b) ⌧MSS = ⌧MDE

(c) ⌧MSS > ⌧MDE

4

o

8avg 0 GETMDER T FmaXl GE ENDE

Tense Zµ f3F 8y2 MDF

0584 TmpE 5YT3

05778

ME 323: Mechanics of Materials

Fall 2020

Homework 11

Due Friday, December 04

2. Cylindrical columns A, B, C and D shown below are made of the same material (Young’s modulusof E) and have the same circular cross-section of radius R. A compressive axial load P is applied toeach column. The critical Euler’s buckling loads PA

cr , PBcr , P

Ccr , and PD

cr for columns A, B, C, and D,respectively, are such that: (2.5 points)

(a) PDcr > PC

cr > PAcr > PB

cr

(b) PDcr > PA

cr > PCcr > PB

cr

(c) PDcr > PB

cr > PCcr > PA

cr

(d) PDcr > PC

cr > PBcr > PA

cr

(e) None of the above.

Provide a written explanation for your answers.

5

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