Azeem Khan

B00230817

Design Analysis 1

Assignment

Bob Bailey

Aim

The aim of this assignment is to produce a design of a structure. The entire structure is required to have a factor of safety of 3.0. In addition to this, unknown dimensions are asked to be calculated so that the structure meets the given conditions.

Calculations supporting the design theory, plus a narrative description of this, will be included in this report.

Introduction

The purpose of this assignment is, ultimately, to assess how the tensile yield strength and the load, can affect the design of a structure i.e. How it affects the size of bolt used, the thickness of materials etc.

Tensile yield strength varies between materials. It is a measure of the force required to pull a structural beam to the point where it breaks. The tensile strength of a particular material is the maximum load of stress that it can take before failure (breaks).

In addition to this, the yield strength is the stress a material can with stand without permanent deformation. It will cause a permanent deformation of 0.2% of the original dimension.

Furthermore, the factor of safety is the ratio of the maximum breaking stress that a structural part can withstand to the estimated maximum stress in ordinary use.

Finally, the impact of the tensile yield strength and the factor of safety on a design structure, as well as the different types of stresses (bearing and shear) are the main principles explored in this assignment.

Calculations

The Question, included in the assignment, is as follows:

A 6m long beam with simply supported ends is required to support a moving point load of 15 tonnes. The section for the beam is shown in Figure 1 and is manufactured from mild steel having a tensile yield strength of 425MN/m2. The beam support arrangement is shown in Figure 2. A side plate is bolted to each side of the beam by two bolts of diameter D1. The bolts have a tensile yield strength of 375MN/m2 and the plates a tensile yield strength of 215MN/m2. The side plates in turn are supported in a cradle as shown via a hinge pin having a diameter D2 having a tensile yield strength of 275 MN/m2. The entire structure is required to have a factor of safety of 3.0.

The design produced should meet the specification given:

Beam Dimensions W, H and T, where H=2*W and T=W20

Bolt Size D1

Side Plate thickness T2 and width W2

Cradle Plate thickness T3 and width W3

Hinge Pin Diameter D2

Take: Bearing Strength = 0.7*Tensile Yield StrengthShear Strength = 0.5*Tensile Yield Strength

The following method is an outline on how to tackle this problem.

1. Calculate y, and in turn the parallel axis theorem allowing us to determine the Ixx. The y and the Ixx are calculated in terms of ‘W’ as no real numbers are given. From the question, the width is taken as W, the height as 2W, and the thickness as W/20.

2. With the aid of FBD, SFD and BMD diagrams, the bending moment acting upon the beam is determined.

3. Calculate the stress acting on the beam from the relationship between the tensile yield stress of the specific material in question and from the factor of safety.

4. From the equation, σ=MYI

; Since Y and Ixx are in terms of W, this

equation can be rearranged to find W, and hence H and T. 5. To find the diameter D1, there are three different stresses acting on the

bolt. The bearing stress on the beam, the bearing stress on the side plate

and the shear stress on the beam. σ= FA

; where the area is equal to π4D2.

Furthermore, the largest value for D is taken to be the diameter as this satisfies the max diameter of all the 3 stresses.

6. Using D1 from above, the thickness of the side plate can be calculated. In addition to this, using this thickness, W2 can be determined.

7. Finally, D2, T3 and W3 can be calculated in that order written as each dimension is dependent on the value before.

2W

W/20

Datum

Shape 1.

Area = w x w20

= w2

20

y = 2W – ( 12 x W20 ) = 2W1

−W40

= 80W40

−W40

=79W40

Ay = W2

20x79W40

= 79W3

40

Shape 2 and 3.

Area = ( 2W1 −W20 )x W20 = ( 40W20 −W

20 ) x W20=39W20 xW20

=39W2

400

y = 12 ( 2W1 −W

20 )=12 ( 40W20 −W20 ) =

12x39W20

= 39W40

Ay = 39W2

400x39W40

=1521W3

16000

∑A = W2

20+ 39W

2

400+ 39W

2

400 = 20W

2

400+ 39W

2

400+ 39W

2

400=98W

2

400

∑Ay = 79W3

800+ 1580W

3

16000+ 1580W

3

16000=1580W

3

16000+ 1521W

3

16000+ 1521W

3

16000=4622W

3

16000=2311W

3

8000

W

Y=∑ Ay∑ A

=¿

2311W8000

3

98W 2

400

= 2311W1960

Parallel axis theorem.

For shape 2 and 3.

H = 2311W1960

−39W40

=2311W1960

−1911W1960

= 400W1960

=10W49

For shape 1

H = 79W40

−2311W1960

=3871W1960

−2311W1960

=1560W1960

=39W49

Ixx=(I ¿¿ xx+A h2)¿ ¿( 112 bd3)+(A ( y )¿¿2)¿

¿ [( 112 xW x ( w20 )3)+(W x

W20x( 39W49 )

2)]+2 [( 112 x W20 x (2W−W20 )

3)+( w20 x (2W−W20 ) x (10W49 )

2)]

¿ [(W12 x W8000

3

)+(W2

20x392W2401

2

)]+2[( W240 x( 40W20 −W20 )

3)+( w20 x ( 40W20 −W20 ) x ( 10W49 )

2)]

= [( W96000

4

+ 1521W4

48020 )]+2[( W240 x 393W8000

3)+( w20 x 39W20 x102W2401

2)]

= [ (0.031685W 4 ) ]+2 [( 59319W1920000

4)+( 3900W960400

4)]

= [ (0.031685W 4 ) ]+2 [ (0.034956W 4 ) ]

= [ (0.031685W 4 ) ]+[ (0.069912W 4 ) ]

= 0.101597W 4

Free Body Diagram

L2FL2

o o

F2F2

Shear Force Diagram

F2F2

o o

−F2

− F2

Bending Moment Diagram

F2xL2= FL4

FL4

o o

From the bending moment diagram, the bending moment acting on this

structure is equal to Force x Length

4

Where theForce=Mass xGravitational Feild Strength=15000 x 9.81=147150N

So, M= F x L4

=147150 x64

=¿ 220725 Nm

The equationσ=MYI

is employed to calculate the width. Y and I are in terms

of W, since W is the only unknown, the equation can be rearranged to determine W. The factor of safety is 3.0. And the tensile yield strength of the beam is 425 N/mm2. This means that the relationship of the direct stress acting upon the beam is equal to

TensileYeild StrenghtFactor of Safety

= 4253

=¿ 141.67 N/mm2 ;

M = 220725 x 103 from before.

σ=MYI ; 141.67 =

220725x 103 x23111960

W

0.101597W 4 ; 141.67 =

220725x 103 x23111960

0.101597W 3

W 3 =220725x 103 x

23111960

0.101597 x141.67

W 3=18081590.18

W 3 = 3√18081590.18

W = 262.47 mm

Hence,

H = 2W = 2 x 262.47 = 524.94 mm

T = W/20 = 262.47/20 = 13.1235 mm

Stress 1. Beam Bearing Stress.

σ = 0.7 x 4253 = 99.167 N/mm2

σ= F4 A =

F

4 xπ4D2 =

F

π D2

Because 4 bolts so 4 x area

99.167=147150π D 2

D2=¿ 147150π x 99.167

= 472.33

D=√472.33=21.73mm

Stress 2. Side Plate Bearing Stress.

σ = 0.7 x 2153 = 50.167 N/mm2

σ= F4 A =

F

4 xπ4D2 =

F

π D2 .

50.167=147150π D2

D2=¿ 147150π x 50.167

= 933.67

D=√933.67=30.56mm

Stress 3. Beam Shear Stress.

τ = 0.5 x 4253

= 70.83 N/mm2

τ= F4 A =

F

4 xπ4D2 =

F

π D2 .

Because 4 bolts so 4 x area

70.83=147150π D2

D2=¿ 147150π x 70.83

= 661.29

D=√661.29=25.72mm

Diameter D1 is taken as 30.56 because it is the largest diameter that satisfies the max diameter of all the 3 stresses acting on the bolt. Since you can’t get/buy a bolt size of 30.56 mm, the bolt size is therefore taken as 32 mm

Thickness T2.

σ= F4 (D 1xT 2) F/4 because force acting on 4

bolts

σ= Side plate bearing stress = 50.167 N/mm2

50.167= 1471504 (32x T 2)

T 2= 1471504 x 32x 50.167 = 22.92 mm

Width W2.

σ= Side plate bearing stress = 50.167 N/mm2

σ= F

2 (W 2−2D 1 ) xT 2

2 side plates = F 2 bolts on 1 side plate so, 2xD1 1 side plate = F/2 W2

Cross Sectional Area + diameter of bolts = W2

2 (W 2−2D1 )T 2 = Fσ

(W 2−2D 1 )T 2 = Fσ x2

W 2−2D 1 = Fσ x2 xT 2

W 2−2D 1 = Fσ x2 xT 2

W 2 = Fσ x2 xT 2 + 2D1

W 2 =

14715050.167 x2 x22.92

+ 2 x 32

W 2 = 63.988 + 64

W 2= 127.99 mm

Diameter D2.

τ = FA

τ = Shear stress of hinge pin = 45.835 N/mm2

τ = Fπ2D2

45.835 = 147150π2D2

D2= 147150π2x 45.835 = 2043.822

D = √2043.822 = 45.21 mm

Can’t get bolt size of 45.21 mm. So, D2 taken as 46 mm

Thickness T3.

σ= 64.167 N/mm2

Where A = 2A because the hinge pin goes through 2 holes,

one at each side of the cradle. So, A = 2 xπ4D2 = π

2D2

σ= F2(D 1x T 3)

T 3= 147150

2x 46 x 64.167 = 24.93 mm

Width W3.

σ= 64.167 N/mm2

σ= F

2 (W 3−D 2 ) x T 3F /2 for oneside of cradle

W3

W 3= Fσ x2 xT 3 + D2

W 3= 14715064.167 x2 x31.89 + 46

W 3= 35.955 + 46

W 3= 81.96 mm

For W3 and T3, assume the tensile strength of the cradle is the same as the hinge pin. This is because the tensile strength for the cradle is not given. Also, the hinge pin goes through the cradle, so the diameter used satisfies the above equations.

F/2 because force acting on 2 sections of the hinge pin, at each side of the cradle

Cross Sectional Area + diameter of bolts = W3

Discussion

To check that the method for determining Y and Ixx are correct; Random values that still meet the specification are used. Where, the height = 2 x W, and Thickness = W/20.

100

200

20

Datum

When W = 100, H = 200 and T = 20;

Shape Area Y AY1 200 x 20 = 2000 190 2000 x 190 = 3800002 180 x 20 = 3600 90 3600 x 90 = 3240003 3600 90 324000∑ 9200 370 1028000

Y=∑ Ay∑ A

¿ 1028009200

=¿ 112

The Y value calculated for the assignment = 1.179W, where W = 100. So, 1.179 x 100 = 117.9. A very small difference in Y values is caused by rounding up numbers.

The parallel axis theorem for shape 1 = 190 – 112 = 78 For shape 2 and 3 = 112 – 90 = 22

Ixx=(I ¿¿ xx+A h2)¿ ¿( 112 bd3)+(Ay )2

¿ [( 112 x100 x203)+(100 x20 x 782 )]+2[( 112 x 20 x1803)+(20 x180 x222 )]= [(66666.67 + 12168000) + 2(9720000 + 1742400)]

= 35159466.67, this value is in terms of the height (because it acts around the x-axis).

My calculated value is in terms of W, where H = 2W.

So, the above Ixx = 35159466.67

2 = 1.7x107

The Ixx value calculated for the assignment is 0.101595W4

= 0.101595 x 1004 = 1.1x107

Again, a difference in values can be due to rounding up numbers

Please turn over

Design

With support from the calculations above, an illustration of the proposed design is as follows:

W

. H

T

Where,

T = 13.13 mm W = 262.50 mm H = 524.94 mm

W2

ØD1

ØD2

W3

Where,

W2 = 127.99 mm W3 = 81.96 mm ØD1 = 32.00 mm ØD2 = 46.00 mm

T2

T3

Conclusion

To conclude this assignment, from the calculations, the structure does indeed have a factor of safety of 3.0 and it does meet the specifications of the structure with the given dimensions. This design is successful and fulfilling to the original aim.

Furthermore, the factor of safety is really important when building a structure, and is very much dependant on the material strength and stresses that act upon it. E.g. When constructing a plane, the margin of safety for an aeroplanes wings is 1.5.

This means that the plane wings can take up to 0.5 times more than the design load. This may not sound so safe compared to the structure of a wardrobe which may have a factor of safety of 5. Theoretically, the factor of safety being 1 is generally sufficient. For an aeroplane, the factor of safety may be low for the following reasons.

Firstly, the design is sufficient in the eyes of the manufacture and in reality this margin is fine.

And the second reason could be to keep manufacturing costs down.

Where,

T2 = 22.92 mm

T3 = 24.93 mm

The smaller the factor of safety, the more likely the cost is lower.