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ENGI 4312Mechanics of Solids I

Prof. Steve Bruneau, EN.4013Ph 737-2119 [email protected] Rizk, Dawood, [email protected]@[email protected]

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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 . . .



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Problem 1

10

marks

Correct answer 5 marksCorrect 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 GPa and 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



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Problem 2

20

marks

Correct answer 10 marksCorrect formulas 10 marks

The torques shown are exerted on pulleys A, B, and C. Knowing that both shafts are solid determine the maximum shearing stress in shaft AB and shaft BC

400 Nm

1200 Nm

800 Nm

Ans = 75.5 MPa and 63.7 MPa

Shear stress=Tau = Tc/J

Torque in AB = 400 Nm

J of AB = ½ pi*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 = ½ pi*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



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Problem 3

20

marks

Correct answers 10 marksCorrect 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 twist between 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



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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 maximum moment for each load case (see data sheet for loading)

0.5m 0.5m

MN 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



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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 side as 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 why not?

Sigma(xx) = M(max)*Cxx / Ixx

20.6*101*10^6/28x10^6 = 73.6 MPa or 10.6 Ksi

32.4*101*10^6/28x10^6 = 115.7 MPa or 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 MPa or 53 Ksi

32.4*64*10^6/3.6x10^6 = 582 MPa or 84.4 Ksi

Ixx = 28x10^6 mm^4

Iyy = 3.6 x10^6 mm^4

64mm

101 mm



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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 MPa while Ixx= 28x10^6mm^4 and y = 101mm

Therefore: 310 = 500*P*101/28x10^6 gives us P of 172 kN for bending

Check direct shear:

area = 3853 mm^2 shear ultimate = 30 KSI = 207 MPa = 799 kN shear resistance – no probs here!

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 kN and 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 mm10.4 mm

91.2 mm

6.4 mm

Potential = 20 extra marks



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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



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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



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Shear Formula

It is important to recall that shear stress is complimentary meaning transverse and longitudinal shear stresses are numerically equal.

The derivation and proof of the shear formula are detailed in Hibbeler Chapter 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

VI

t

''''

AydAyQA

== ∫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 centroid of A’, measured from the neutral axis

'y

Transverse Shear

Hibbeler

Chapter 7



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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.

ItVQ

=τ

Shear Stresses in Beams

Applying the shear formula for common beam cross-sectional situations:

Rectangular: ( )[ ]( )[ ]bbh

byhV

ItVQ

3

22

12/1

4/21

−⎟⎠⎞

⎜⎝⎛

==τ

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

2

3 46 yhbhVτ

AV5.1max =τ

AV

avg =τMaximum 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)



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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

Wide Flange Beams:

A wide flange beam consists of two flanges and a web. An analysis of the shear in a wide flange beam results in the illustration below:

Parabolic

Jump due to the smaller “t” in the shear formula

Transverse Shear

Hibbeler

Chapter 7



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In summary the important points to remember about shear stresses in beams are:

Transverse Shear

Hibbeler

Chapter 7



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Transverse Shear

Hibbeler

Chapter 7



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Examples:

Transverse Shear

Hibbeler

Chapter 7

ItVQ

=τ

''''

AydAyQA

== ∫

Key Formulas for Solution:



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Transverse Shear

Hibbeler

Chapter 7

''''

AydAyQA

== ∫



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Transverse Shear

Hibbeler

Chapter 7

''''

AydAyQA

== ∫



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Transverse Shear

Hibbeler

Chapter 7

ItVQ

=τ



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Transverse Shear

Hibbeler

Chapter 7

ItVQ

=τ



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Examples:

Transverse Shear

Hibbeler

Chapter 7

ItVQ

=τ ''''

AydAyQA

== ∫



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Transverse Shear

Hibbeler

Chapter 7



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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 =

''''

AydAyQA

== ∫

Where A’ is the top or bottom portion of the member’s cross sectional area, defined from the section where t is measured, and ybar is the distance to the centroid of A’, measured from the neutral axis



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Transverse Shear

Hibbeler

Chapter 7

Shear Flow in Built up MembersI couldn’t say it any better than Hibbeler so read the following:



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Transverse Shear

Hibbeler

Chapter 7

Examples of Shear Flow

433 231.23)75.4)(5.5(121)25.5)(6(

121 inI =−=

375.3)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

ItVQB

allow ==τ

kiplbV 24.11239 ==

N A

5.5 in0.25 in0.25 in

0.25 in

5 in2.5 in

2.5 in

a

b



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Transverse Shear

Hibbeler

Chapter 7

Examples of Shear Flow

433 231.23)75.4)(5.5(121)25.5)(6(

121 inI =−=

375.3)25.0)(6(5.2'' inAyQb ===

psiIt

VQBb 646

)25.0)(2(231.23)75.3)(10(2 3

===τ

N A

5.5 in0.25 in0.25 in

0.25 in

5 in2.5 in

2.5 in

a

b

34375.3)25.0)(5.5(5.2 inQa ==

psiIt

VQaa 592

)25.0)(2(231.23)4375.3)(10(2 3

===τ

0.25 in



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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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