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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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In summary the important points to remember about shear stressesin

beams are:

Transverse Shear

Hibbeler

Chapter 7

ENGI 4312

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Transverse Shear

Hibbeler

Chapter 7

ENGI 4312

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Examples:

Transverse Shear

Hibbeler

Chapter 7

It

VQ

=

τ

'''

'

AydAyQ

A

==

∫

Key Formulas for Solution:

ENGI 4312

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Transverse Shear

Hibbeler

Chapter 7

'''

'

AydAyQ

A

==

∫

ENGI 4312

Mechanics of Solids I Prof. Steve Bruneau, EN.4013

Ph 737-2119

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Transverse Shear

Hibbeler

Chapter 7

'''

'

AydAyQ

A

==

∫

ENGI 4312

Mechanics of Solids I Prof. Steve Bruneau, EN.4013

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Transverse Shear

Hibbeler

Chapter 7

It

VQ

=

τ

ENGI 4312

Mechanics of Solids I Prof. Steve Bruneau, EN.4013

Ph 737-2119

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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

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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

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Transverse Shear

Hibbeler

Chapter 7

ENGI 4312

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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

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Transverse Shear

Hibbeler

Chapter 7

Shear Flow in Built up Members

I couldn’t say it any better than Hibbelerso read the following:

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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

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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

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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

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