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HNC/HND in Mechanical EngineeringBTEC Diploma in Engineering

Unit 2Engineering science

Behavioural characteristics of elements ofstatic engineering systems

Created by:Gabriel Stancu

HND Mechanical EngineeringYEAR 1

2013-10-31

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CONTENTS

Summary...................................................................page 3

Introduction ……….....................................................page 3

Lab Report #1 Shear force and Bending moment…………………………………………………….............page 4

Lab Report #2 Slenderness Ratio and Critical Buckling

Load ...........................................................page 5

Lab Report #3 Shear stress distribution and angular

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deflection…………………………………..............................page 6

Conclusion………........................................................page 7

References………........................................................page 8

Summary

The purpose of this assignment is to demonstrate the knowledge of basic engineering problems involving columns, bending and torsion, and to investigate the loading of engineering components focussing on the loading of columns, torsional effects on shafts and bending of beams.

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Introduction

The first part of the assignment is a laboratory test to determine the maximum stress and the radius of curvature due the bending for a simply supported ABS beam.

The second part of the assignment is to determine the slenderness ratio and select a suitable standard rolled steel for the column

The last part of the assignment is to determine the angular deflection and plot the distribution of the shear stress in the shaft.

TASK1Lab report #1

Date:

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10/11/2013

Objectives:

Determine distribution of shear force, bending moment and stress and indicating the max stress and radius of gyration of the beam selected

Apparatus:

- Two sets of weights: 100grams and 50 grams- Support frame- Dial Gauge- Caliper- Ruler- Balance- Pen- Record sheet

Test specimen:

- ABS Plastic beam H shape

Procedure:

The methodology involved in this report is as follow (Figure1):

- The beam was measured with the ruler and weighed on the balance.

- The thickness of the beam was measured with thecaliper

- The data was recorded on the record sheet- The ABS beam was placed on the support frame.

The beam was simply supported on 2 points at 600mm

- The dial gauge was reset to zero- The 100grams weight was placed at 100mm from

the left support

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- The 50grams weight was placed at 100mm from theright support.

- The reading of the dial gauge was recorded on arecord sheet

Figure 1: ABS beam simply supported

Records:

- ABS beam total length: 650mm = 650*10-3 m- ABS beam length supported = 600mm =

600*10-3 m- ABS beam weight: 32.54grams- Depth = 9.8mm = 9.8*10-3 m- Weight1 = 100g- Weight2 = 50g- Modulus of elasticity ABS beam = 2.8*109 N/m2

- Second moment of area ABS beam = 0.5*10-9

- Dial gauge = 2.2mm = 2.2*10-3 m

Calculation:

To calculate the maximum stress force and the radius of curvature, the following equations should be used:

- CW = ACW- UpF = DF- M/I = τ/y = E/R- I = (BD3)/12- F = (weight/1000) * 9.8- UDL = [(weight of the beam/1000) * 9.8]/L- y = D/2

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Where,

CW= clockwise forces (N), ACW= anticlockwise forces (N), UpF= upper forces (N), DF=down forces (N), M= applied moment (N/m), I = moment of inertia of beam (m4), τ= Stress (N/m2), y= neutral axis (m), E= strain , R= radius of deflection , F= force (N) , UDL= uniformly distributed load (N) , L= total length(m)

1.CW = ACW and UpF = DF (Figure 2)

Figure 2: Simply supported beam scheme

F1 = (weight1/1000)*standard gravity

∴ F1 = (100/1000)*9.8

∴ F1 = 980*10-3 N

F2 = (weight2/1000)*standard gravity

∴ F2 = (50/1000)*9.8

∴ F2 = 490*10-3 N

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UDL = [(weight of beam/1000)*standard gravity]/Length

∴ UDL= [(32.54/1000)*9.8]/650*10-3

∴ UDL=490*10-3 N/m

2.CwF = ACwF

∴ (0.98*0.1) + (0.49 *0.5) + (0.49*0.6*0.3) = R2*0.6

∴ 0.049 + 0.49 + 0.09 = R2*0.6

∴ R2 = 719*10-3 N

UpF = DF

∴ R1 + 0.719 = 0.98 + 0.49 + (0.49*0.6)

∴ R1 = 1.045 N

3.Shear force and Moment Diagram (Figure 3)

Shear Force

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Figure 3: Shear force and bending moment diagram

4.Max Bending moment, Bending stress and radius of gyration

@0.1m = (1.045*0.1)-(0.49*0.1*0.05) = 1.102N/m

@0.12m = (1.045*0.12)-(0.98*0.02)-(0.49*0.12*0.06) = 0.10227N/m

@0.13m = (1.045*0.13)-(0.98*0.03)-(0.49*0.13*0.065) =0.1023N/m

@0.14m = (1.045*0.14)-(0.98*0.04)-(0.49*0.14*0.07) = 0.10229N/m

@0.16m = (1.045*0.16)-(0.98*0.06)-(0.49*0.16*0.08) = 0.10212N/m

@0.18m = (1.045*0.18)-(0.98*0.08)-(0.49*0.18*0.09) = 0.1017N/m

@0.2m = (1.045*0.2)-(0.98*0.1)-(0.49*0.2*0.1) = 1.1012N/m

Bending Moment M = 102.3*10-3 N /m

M/I = τ/y

M = 102.3*10-3 N/m

Bending Moment

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I = 0.5*10-9 m4

y= 4.75mm = 4.75*10-3 m

∴ τ = (102.3*10-3*4.75*10-3)/0.5*10-9

Bending Stress τ = 971.85*103 N/m2

M/I = E/R

M = 102.3*10-3 N/m

I = 0.5*10-9 m4

E = 2.8*109 N/m2

∴ R = (2.8*109*0.5*10-9)/102.3*10-3

Radius of gyration R = 13.68 m

Conclusion:

The bending moment acting at a point on a beam is the resultant due to all the forces acting the one side of the point.

The relation between shear force and bending moment is illustrated in figure 3/ page 7: - The shear force is maximum at support and zero at centre of bending; the bending moment is zero at support and maximum at centre bending - The bending moment is zero at support and maximum at centre of bending

τ = MY/I or τ = Ey/R

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Analysing both equations result that the stress in a beam depends on the bending moment and sothe maximum stress will occur where the bending moment is a maximum along the length of the beam, it also indicates that stress is related to distance y from the neutral axis so it varies from zero to maximum at the top or at the bottom of the section. One edge of the beam will be in maximum tension and the other in the maximum compression. So, in reality,bending stresses are tensile or compressive stresses in the beam! A simply-supported beam always has tensile stresses at the bottom of the beam and compressive stresses at the top of the beam. Figure 4

Figure 4: Distribution of shear and stress

When a simply supported beam is loaded in flexure, the top side is in compression, and the bottom side is in tension. If the flexural load increases to a critical limit, the beam will fail dueto lateral buckling of the compression flange. The slenderness ratio is calculated by the formula:

S.R. = Le/r

Stress

Strain

CompressionTension y

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Where: S.R – slenderness ratio, Le – effective length(m), r – radius of gyration (m)The radius of gyration is calculated by the formula:

r = √ I/A

Where r – radius of gyration (m), I = second moment of area (m4), Cross section of area.For the test specimen - plastic ABS beam H section:

I=0.5*10-9 m4 and A=41.9*10-6 m2

∴ r= √ 0.5*10-9/41.9*10-6

∴ r=3.45*10-3 m∴ S.R. = 0.6/3.45*10-3

∴ S.R. = 173.9 N

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Lab report #2

Date: 10/11/2013

Objectives:

Assuming that the beam from Lab report #1 is transposed to a scale 1:20 and is used as a column, specify if whether the column is treated as a long oras a short column, under axial compressive load.

Test specimen:

- Steel column

LOAD

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Figure 5

Records:

Column Length = 0.65*25 = 16.25m

Column depth = 9.5*103 * 25 = 237.5*103 m

According the ‘’Table for universal columns with H section’’ the nearest depth section to 237.5mm is 254 mm – designation for the standard rolled steel H shape column 254x254x73. (Appendix 1)

I = 11407 cm4 = 114.07*10-6 m4

A = 93.1 cm2 = 931*10-6 m2

E = 200 GPa = 200*109 N/m (Wikipedia - Young's modulus)

Where,

I – second moment of area (m4), A – section area(m2), E – modulus of elasticity steel (N/m2)

Calculation:

1.Slenderness ratio.

STEELCOLUM

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S.R. = Le/r

Where: S.R – slenderness ratio, Le – effective length (m), r – radius of gyration (m)

The radius of gyration is calculated by the formula:

r = √ I/A

Where: r – Radius of gyration (m), I = second moment of

area (m4), A - Cross section of area (m2)∴ r = √114.07*10-6/931*10-6

∴ r = 0.35 m

∴ S.R. = 16.25/0.35∴ S.R. = 46.42

2.The Critical Buckling load.

Pcr = π2 * E*I/L2 (Euler equation)

Where:Pcr – critical buckling load (N), E - modulus of elasticity (N/m2), I – secondmoment of area (m4), L – effective length (m)

∴ Pcr = π2 * 200*109 * 114.07*10-6 / (16.25)2

∴ Pcr = 852.7*103 N

Conclusion:

A column can either fail due to compression or buckling when is loaded. The critical buckling load

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is the critical or maximum axial load on the column just before it begins to buckle.

The slenderness ratio is used to help determine when the Euler's column buckling equation is valid. If the column is short (low slenderness ratio) then the column will be crushed before it buckles. 

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Lab report #3

Date:

10/11/2013

Objectives:

Determine distribution of shear stress and the angular deflection due to torsion in circular shafts at 10 º

Apparatus:

- Five sets of weights100grams, 150 grams, 200 grams and 250grams

- Support frame - Dial scale ( Figure 5)- Caliper- Ruler- Pen- Record sheet

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Figure 5

Test specimen:

- Steel Rod

Procedure:

The methodology involved in this report is as follow:

- The steel rod length was measured with the ruler

- The steel rod diameter was measured with the caliper

- The data measurements was recorded on the record sheet

- The steel rod was fixed on the support frame , by locking the chucks

- The dial scale was set to zero- The distance to pivot point was measured with

the ruler- On the dial scale different weights were placed

to observe angle of deflection due the torsion of the shaft : 100 grams, 150 grams, 200 grams and 250 grams

- The data measurement was recorded on the recordsheet.

- After analyzing the data recorded the Shear stress and the Angle of deflection was calculated to a force applied F= 2.15N to an angle of 10o. (Figure 6)

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Figure 6

Records:

- Steel shaft length = 450mm = 450*10-3 m

- Steel shaft diameter = 3mm = 3*10-3 m

- Steel shaft radius = 1.5mm = 1.5*10-3 m

- Distance to pivot point = 100mm = 100*10-3 m

- Modulus of rigidity of steel = 82GN/m2 = 82*109N/m2

Calculation:

To calculate the maximum stress force and the radius of curvature, the following equations should be used:

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- T/J = τ/r = (G*θ)/L- F = (weight * 9.8)/1000- J = πd4/32- T = F * D

Where,

T = Applied Torque; (Nm), J = Polar Second Momentof Area; (m2), τ = Shear Stress (N/m2), r = Radius (m), G = Modulus of Rigidity; (N / m2), θ = Angle of Twist (over length l); (radians), L= Shaft Length (m), d = diameter of shaft (m), D = distance to pivotpoint

F = weight*9.8/1000

∴ F1 = 0.98N at 5 º

∴ F2 = 1.47N at 7.5 º

∴ F3 = 1.96N at 9.5 º

∴ F4 = 2.54N at 11 º

Analysing the chart (Figure 6/pag 10) can be observed that F5 = 2.15N at 10 º

J = πd4/32

∴ J = π (3*10-3)4/32

∴ J = 7.95*10-12 m4

Torque at 10 º:

F = 2.54 N

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D = 100*10-3 m

T4 = F * D

∴ T4 = 2.54 * 100*10-3

∴ T4 = 254*10-3 N

T/J = τ/r

T = 254*10-3 N

J = 7.95*10-12 m4

r = 1.5*10-3 m

∴ τ = (T * r)/ J

∴ Shear stress τ = 47.9*106 N/m2

T/J = (G*θ)/L

T = 254*10-3 N

J = 7.95*10-12 m4

G = 82*109 N

L = 450*10-3 m

∴ θ = (T*L)/(G*J)

∴ Angle of deflection θ = 175.3*10-3 radians

Conclusion:

Torsion occurs when any shafts is subject to a torque. This is true whether the shaft is rotating orstationary. The torque makes the shaft Twist and one

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end rotate relative to other inducing shear stress onany cross section. Failure might occur due the shear alone or because the shear is accompanied by stretching or bending.

The diagram nr 1 shows the steel rod fixed at oneend and twisted at the other to the action of a torque T = 254*10-3 N. The radius of the shaft is r =1.5*10-3 m and the length is L = 450*10-3 m. It can beobserved that due the torque applied the shaft twisted at an angle θ = 175.3*10-3 radians. Convertingthe radian in degrees resulted an angle θ = 10o.

Diagram Nr1 : Angle of Twist

The distribution of the shear stress on a cross section is shown graphically in diagram nr.2 as varying linearly with the distance from the axis of rotation of the shaft with the maximum stress.

Diagram Nr 2: Shear stress distribution

Radius = 1.5*10-3

Angle of twist = 10o

Torque = 254*10-3

N

Length = 450*10-3 m

r

τ

r0

τ max=Tr0/J

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References:

Stresses in beams due to bending. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/statics/beams/beam%20tut1.pdf. [Accessed 17 November 2013].

Tutorial 3- Torsion. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/c103/t3.pdf. [Accessed 17 November 2013].

Bending Moments. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/engineering%20science%20h1/outcome%201%20t3.pdf. [Accessed 17 November 2013].

Columns. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/Eng%20Science%20HNC%20Unit%202/OUTCOME1/OUTCOME1%20T4.pdf. [Accessed 17 November 2013].

Stresses in beams due to bending. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/Eng%20Science%20HNC%20Unit%202/OUTCOME1/OUTCOME1%20T1.pdf. [Accessed 17 November 2013].

Roy Beardmore. (25/01/2013). Elastic bending. Available: http://www.roymech.co.uk/Useful_Tables/Beams/Beam_theory.html. Last accessed 17/11/2013.

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The deflection of beams. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/statics/beams/beam%20tut3.pdf. [Accessed 17 November 2013].

B2Bmetal.eu . (2013). European standard wide flange Hsteel beams specifications. Available: http://www.b2bmetal.eu/heb-sections-specification. Last accessed 24/11/2013.

engineeringtoolbox. (2013). Young Modulus . Available: http://www.engineeringtoolbox.com/young-modulus-d_417.html#.UpFvOtK8DIY. Last accessed 24/11/2013.

Tutorial 4 - Columns. 2013. . [ONLINE] Available at: http://www.freestudy.co.uk/engineering%20science%20h1/outcome%201%20t4.pdf. [Accessed 24 November 2013].