sm lab report 1 senior

Curtin University

Department of Civil Engineering

Structural Mechanics 262

Laboratory Report Attention: Prof. Marcus Lee

Laboratory 1: Shear Centre

Lab 1: p1

Structural Mechanics 262 - Laboratory 1: Shear Centre

Student Number / Name:

Jeffrey Lam Khai Yue 7E1B9006 / 15834566

Kong Shin Hwa 7E1A8399 / 15628277

Lu Yuan Jie 7E1A8597 / 15605788

Rebecca Lim Jye 7E1A8334 / 15587213

Laboratory Group Number: 9

Attendance Date: 7

th October 2013

Report Due Date: 25

th October 2013

Marking Key:

Presentation, Layout and

Compliance with report

requirements

(2)

Calculation/Discussion

(8)

Total Marks

(10)

Reduction of mark for

late submission

(20% per day)

Final Recorded Mark

(10)

Lab 1: p2


Table of Contents

1.0 Introduction ................................................................................................................. 3

1.1 Background Information ......................................................................................... 3

1.2 Objectives ................................................................................................................ 4

2.0 Procedures ................................................................................................................... 5

3.0 Results ......................................................................................................................... 6

3.1 Experimental Shear Centre ..................................................................................... 6

3.1.1 Section 1: C-Channel Section .............................................................................. 6

3.1.2 Section 2: Unsymmetrical I-Section .................................................................... 7

3.1.3 Section 3: Extended C-Channel Section .............................................................. 8

3.2 Theoretical Shear Centre ......................................................................................... 9

3.2.1 Section 1: C-channel Section ............................................................................... 9

3.2.2 Section 2: Unsymmetrical I-Section .................................................................. 10

3.2.3 Section 3: Extended C-Channel Section ............................................................ 11

3.3 Summary of Results .............................................................................................. 12

4.0 Discussion ................................................................................................................. 13

5.0 Conclusion ................................................................................................................ 17

6.0 References ................................................................................................................. 18

Appendix .............................................................................................................................. 19

Lab 1: p3


1.0 Introduction

1.1 Background Information

The shear centre is the point where a force can be applied to cause the section to

bend without any twisting or torsion. Thus on symmetrical sections the shear centre is

usually located on the centre of twist of that section, which coincides with the centroid of

the beam as the shear centre can be found at the intersection of the X and Y axis of

symmetry.

Figure 1. Shear Centre (P.P.Benham, 2012)

However, when it comes to unsymmetrical section (i.e. equal angle, unequal angle,

or channel section), the position of shear centre would not coincide with the centroid of the

section as a set of forces must be in equilibrium with the applied vertical force so that it

will not be twisted. When a force is applied at any point on a section that are symmetrical

in only one axis, bending of the section will occur and eventually leads to torsion. The

shear flow distribution in the flanges and web produces resultant forces in each member.

Thus, a resultant moment will be obtained when moments are summed about a point. This

resultant moment is known as a torque or couple which causes the twisting of the member.

To avoid twisting of the member, the load applied needs to be located in certain

point with eccentricity, e from the centroid of the section. Otherwise in a simpler form of

explanation, shear centre is defined as the point on the beam section where load is applied

and no twisting is produced (P.P.Benham, 2012).

Lab 1: p4


In order to determine the position of shear centre, formulas can be derived from the

shear flow theory. Thus formulas for section C-Channel section, Unsymmetrical I-Section

and Extended C-Channel section are derived for the ease of calculation. The derivation can

be referred to Appendix A, Appendix B and Appendix C respectively. The derived

formulas are as shown:

� − �� : � = ��

�� − !��"�: � = �� #�$� − ��%

&'��(�( ) − )*�� !��"�: � = �� +�� + �� − ��-�� .

1.2 Objectives

The objective of this experiment is to experimentally determine the location of the

shear centre of various sections (C-Channel section, Unsymmetrical I-Section and

Extended C-Channel section) and compare these values with the theoretical results which

can be determined by calculating the theoretical location of the shear centre of each of the

sections.

Lab 1: p5


2.0 Procedures

1. The dimension of the section were measured and recorded on the lab sheets (on

section diagrams).

2. The two dial gauge plungers were set to bear on the machined surfaces of the

attached plate when no load are on the specimen. The dial gauge datum was then

recorded.

3. The specimen was positioned initially at the outer-most notch from the web and

was then loaded with 4kg weight. The left- hand (LH) and right – hand (RH) dial

gauge reading were both recorded.

4. The load was moved towards the next notch of the web and the dial gauge

readings were recorded again. The loads were to be shifted towards the next

notch one at a time and the readings of all the positions situated 20mm apart were

recorded.

5. Step 1 to step 4 were carried out for C-Channel section, Unsymmetrical I-Section

and Extended C-Channel section.

Lab 1: p6


3.0 Results

3.1 Experimental Shear Centre

3.1.1 Section 1: C-Channel Section

Table 1. Experimental results of C-Channel section

Position

LH dial gauge: RH dial gauge:

Reading

(mm)

Deflection

(mm)

Reading

(mm)

Deflection

(mm)

Datum 7.00 0.00 9.00 0.00

-100 6.61 -0.39 9.43 0.43

-80 6.70 -0.30 9.31 0.31

-60 6.77 -0.23 9.21 0.21

-40 6.84 -0.16 9.11 0.11

-20 6.90 -0.10 9.02 0.02

0 6.99 -0.01 8.90 -0.10

20 7.13 0.13 8.76 -0.24

40 7.26 0.26 8.62 -0.38

60 7.43 0.43 8.29 -0.71

80 7.54 0.54 8.16 -0.84

100 7.66 0.66 8.04 -0.96

Figure 2. Graph of deflection against load position for C-Channel Section

Figure 2 shows the graph of deflection against load position for C-Channel section

which is plotted based on the experimental result. Based on the graph, the intersection

point between the two straight lines indicates the experimental shear centre of C-Channel

section which is 22.50 mm to the left side of the section.

y = 0.005x + 0.075

y = -0.007x - 0.195

-1.20

-1.00

-0.80

-0.60

-0.40

-0.20

0.00

0.20

0.40

0.60

0.80

-110-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110

Loa

d P

osi

tio

n (

mm

)

Deflection (mm)

Graph of Deflection against Load Position

LH Deflection

RH Deflection

Lab 1: p7


3.1.2 Section 2: Unsymmetrical I-Section

Table 2. Experimental results of Unsymmetrical I-Section

wePosition


Reading

(mm)

Deflection

(mm)

Reading

(mm)

Deflection

(mm)

Datum 7.00 0.00 8.00 0.00

-100 6.46 -0.54 8.46 0.46

-80 6.54 -0.46 8.31 0.31

-60 6.64 -0.36 8.18 0.18

-40 6.74 -0.26 8.08 0.08

-20 6.84 -0.16 7.98 -0.02

0 6.94 -0.06 7.88 -0.12

20 7.05 0.05 7.76 -0.24

40 7.15 0.15 7.65 -0.35

60 7.26 0.26 7.54 -0.46

80 7.38 0.38 7.43 -0.57

100 7.48 0.48 7.33 -0.67

Figure 3. Graph of deflection against load position for Unsymmetrical I-Section

Figure 3 shows the graph of deflection against load position for Unsymmetrical

I-Section which is plotted based on the experimental result. Based on the graph, the

intersection point between the two straight lines indicates the experimental shear centre of

Unsymmetrical I-Section which is 8.00 mm to the left side of the section.

y = 0.005x - 0.047

y = -0.005x - 0.127

-0.80

-0.70

-0.60

-0.50

-0.40

-0.30

-0.20

-0.10

0.00

0.10

0.20

0.30

0.40

0.50

0.60

-110-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110

Loa

d P

osi

tio

n (

mm

)

Deflection (mm)


LH Deflection

RH Deflection

Lab 1: p8


3.1.3 Section 3: Extended C-Channel Section

Table 3. Experimental results of extended C-Channel section

Position


Reading

(mm)

Deflection

(mm)

Reading

(mm)

Deflection

(mm)

Datum 2.00 0.00 4.00 0.00

-100 1.88 -0.12 4.09 0.09

-80 1.90 -0.10 4.05 0.05

-60 1.92 -0.08 4.02 0.02

-40 1.94 -0.06 3.99 -0.01

-20 1.96 -0.04 3.96 -0.04

0 1.98 -0.02 3.92 -0.08

20 2.00 0.00 3.89 -0.11

40 2.02 0.02 3.86 -0.14

60 2.04 0.04 3.82 -0.18

80 2.07 0.07 3.79 -0.21

100 2.09 0.09 3.76 -0.24

Figure 4. Graph of deflection against load position for Extended C-Channel Section

Figure 4 shows the graph of deflection against load position for Extended C-

Channel Section which is plotted based on the experimental result. Based on the graph, the

intersection point between the two straight lines indicates the experimental shear centre of

Extended C-Channel section which is 29.50 mm to the left side of the section.

y = 0.001x - 0.018

y = -0.001x - 0.077

-0.30

-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

-110-100 -90 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70 80 90 100 110

Loa

d P

osi

tio

n (

mm

)

Deflection (mm)


LH Deflection

RH Deflection


3.2 Theoretical Shear Centre

3.2.1 Section 1: C-channel Section

Figure

The shear center of C-Channel section is calc

Based on the measured dimension of C

b = 48.60 mm, h = 99.13 mm,

/01234567 89 = :;<=> ??

@AA = B +C6DEF + GHI.

= JE<:: K LM<:NDEF O + J>L

= >LP;==<N= ??Q

0 = >N<P;I K LL<E=I K E<::> K >LP;==<N=

= EN<E= ??

Therefore, shear centre of C

the section along the axis of symmetry.

Please refer to Appendix A for the derivation of the equation of

section.


Theoretical Shear Centre

channel Section

Figure 5. Dimensions for C-Channel section

Channel section is calculated using the equation below:

� = ��

dimension of C-Channel section,

48.60 mm, h = 99.13 mm, t = 1.55 mm

.

J>L<=M K E<::DEF + R>L<=M K E<::SRT:;<=> − ;

::

shear centre of C-Channel section is located at 18.13 mm to the left hand side

along the axis of symmetry.

Please refer to Appendix A for the derivation of the equation of shear for C

Lab 1: p9

ulated using the equation below:

;<MNTISO K F

to the left hand side of

shear for C-Channel


3.2.2 Section 2: Unsymmetrical I

Figure

The shear center of Unsymmetrical I

Based on the measured dimension of Unsymmetrical I

b1 = 39.55 mm, b2 = 19.24mm, h = 100.24 mm, t = 1.24 mm

/01234567 89 = :;<M> ??


= JE<F> K LN<MPDEF O + J:N

= >PP:L><L; ??Q

0 = E;;<F>I K E<F> K R=L<> K >PP:L>

= M<LM ??

Therefore, shear centre of unsymmetrical I

side of the section along the axis of symmetry.

Please refer to Appendix A for the derivation of the equation of shear for

I-Section.


Section 2: Unsymmetrical I-Section

Figure 6. Dimensions for Unsymmetrical I-Section

The shear center of Unsymmetrical I-Beam is calculated using the equation bel

� = ��#�$� − ��%��

Based on the measured dimension of Unsymmetrical I-Section,

= 19.24mm, h = 100.24 mm, t = 1.24 mm

.

J:N<N; K E<F>DEF + R:N<N; K E<F>SRT:;<M> − ;

R <::I − EL<F>IS>PP:L><L;

Therefore, shear centre of unsymmetrical I-Section is located at 7.97 mm



Lab 1: p10

Beam is calculated using the equation below:

;<PFTISO K F

Section is located at 7.97 mm to the left hand

Please refer to Appendix A for the derivation of the equation of shear for Unsymmetrical


3.2.3 Section 3: Extended C

Figure

The shear center of Extended C

Based on the measured dimension of

b = 49.75 mm, h = 99.38 mm

/01234567 89 = :;<== ??


= JE<FM K LN<EEDEF O + +

+ +JE<FM K F><;>DEF +

= :EFLP:<;P ??Q

0 = >L<M: K E<FM +>L<M: K

= FL<FF ??

Therefore, shear centre of extended C

hand side of the section along the axis of symmetry.


C-Channel section.


Section 3: Extended C-Channel Section

Figure 7. Dimensions for Extended C-Channel section

ear center of Extended C-Channel is calculated using the equation below:

� = �� +�� + �� − ��-�� .

Based on the measured dimension of Extended C-Channel section,

b = 49.75 mm, h = 99.38 mm, t = 1.27 mm, a = 24.68 mm

.

+J:E<;F K E<FMDEF + R:E<;F K E<FMSRT:;<== −

+ RE<FM K F><;>SRT:;<== − E=<FLTISO K F.

K LL<=NI + F K F><PN K LL<=NI − F K F><PN> K :EFLP:<;P

shear centre of extended C-Channel section is located at 29.22 mm



Lab 1: p11

Channel is calculated using the equation below:

− ;<P>TISO K F.

.

PND.

Channel section is located at 29.22 mm to the left

Please refer to Appendix A for the derivation of the equation of shear for Extended

Lab 1: p12


3.3 Summary of Results

Table 4. Summary of results for experimental shear centre and theoretical shear centre

Sections

Experimental

Shear Centre

(mm)

Theoretical

Shear Centre

(mm)

Section 1

22.50 18.13

Section 2

8.00 7.97

Section 3

29.50 29.22

Table 4 above shows the summary of results for the experimental shear centre and

theoretical shear centre. The comparison between these two results is further explained in

the discussion.


4.0 Discussion

In this laboratory report, both the experimental and theoretical shear centre is

obtained for three different sections. The theoretical shear centre for three different

sections was calculated using their respective shear centre’s f

experimental shear centre, a graph of left hand and right hand deflections against the load

position was plotted. The intersection point between the two best fit lines indicates the load

position where the channel frame is experiencing e

intersection is defined as the shear centre.

Comparison of theoretical and experimental results

Figure

The internal shear flow in a C

in the C-Channel section flows from point A to point E. If the load is applied not on the

shear centre with the eccentricity, e, the C

the internal shear flow. This happens as the

the equilibrium condition of the C

From the experimental results obtained from the graph in

is located 22.50 mm to the left hand side of the section along the axi

other hand, the theoretical shear centre calculated is 18.13

section along the axis of symmetry.

shear centre found, the percentage error is calculated

U03V012WX0 Y3343 = ZU03V012WX0 Y3343 = Z

= F>

Therefore, the percentage er




sections was calculated using their respective shear centre’s formula. As for the



position where the channel frame is experiencing equilibrium. Thus, the point of

intersection is defined as the shear centre.

theoretical and experimental results

Figure 8. Shear flow in C-Channel Section

he internal shear flow in a C–Channel section is shown in figure 8

hannel section flows from point A to point E. If the load is applied not on the

ith the eccentricity, e, the C-Channel section tends to twist to the right due to

the internal shear flow. This happens as there is no force applied to balance and maintain

quilibrium condition of the C-Channel section.

From the experimental results obtained from the graph in figure 2

mm to the left hand side of the section along the axis of symmetry. On the

other hand, the theoretical shear centre calculated is 18.13 mm to the left hand side

the axis of symmetry. Comparing both these experimental and theoretical

shear centre found, the percentage error is calculated by using the equation below:

Z2H043025VW[ \W[]0 − 0^_035?012W[ \W[]02H043025VW[ \W[]0

ZEN<E= − FF<:;EN<E= Z

F><E; `

, the percentage error for the C-Channel Section is 24.10 %.

Lab 1: p13



ormula. As for the



quilibrium. Thus, the point of

figure 8. The shear flow

hannel section flows from point A to point E. If the load is applied not on the

hannel section tends to twist to the right due to

re is no force applied to balance and maintain

figure 2, the shear centre

s of symmetry. On the

mm to the left hand side of the

Comparing both these experimental and theoretical

by using the equation below:

\W[]0Z a RES


Figure

Figure 9 shows the s

C-Channel section, if the load is

right hand side due to the internal shear flow of the section.

From the graph in figure 3

is 8.00 mm to the left hand side

theoretical shear centre, it is calculated to be 7.97

along the axis of symmetry

centre found, the percentage error

U03V012WX0

Therefore, the percentage er


Figure 9. Shear flow in Unsymmetrical I-Section

shows the shear flow in an unsymmetrical I–Section. Similarly to the

if the load is not applied at the shear centre, the section will twist to the

right hand side due to the internal shear flow of the section.

figure 3, it is shown that the experimental shear centre obtained

mm to the left hand side of the section along the axis of symmetry

shear centre, it is calculated to be 7.97 mm to the left hand side

along the axis of symmetry. Comparing both these experimental and theoretical shear

centre found, the percentage error is calculated by using equation (1) as shown below.

U03V012WX0 Y3343 = ZM<LM − N<;;M<LM Z

= ;<=N `

, the percentage error for the unsymmetrical I-Section is 0.38 %.

Lab 1: p14

Section. Similarly to the

not applied at the shear centre, the section will twist to the

is shown that the experimental shear centre obtained

along the axis of symmetry. As for the

mm to the left hand side of the section

Comparing both these experimental and theoretical shear

is calculated by using equation (1) as shown below.

%.


Figure

According to figure

from point A to F. This internal shear flow will cause the whole section to twist to the right

hand side if no force is applied on the shear centre to balance the section.

Based on the graph in figure

side of the section along the symmetrical axis and the theoretical shear centre found is

29.22 mm to the left hand side of the section which also lies on the axis of symmetry.

Comparing both these experi


U03V012WX0

Therefore, the percentage error obtained for the extended C

Taking into account the theoretical shea

section (section 1), unsymmetrical I

section (section 3), section 2 have the smallest value. Hibbeler (2011, 394) clai

location of the shear centre is a function of the geometry of the cross section. This is due to

the cross sectional area of section 2, having almost two axes of symmetry and thus

producing a smaller value of eccentricity

the section.


Figure 10. Shear flow in Extended C-Channel Section

According to figure 10, the shear flow direction in an extended C



graph in figure 4, the shear centre is located 29.50 mm to the left hand


mm to the left hand side of the section which also lies on the axis of symmetry.

Comparing both these experimental and theoretical shear centre found, the percentage error


U03V012WX0 Y3343 = ZFL<FF − FL<:;FL<FF Z

= ;<LP `

Therefore, the percentage error obtained for the extended C-channel section i

Taking into account the theoretical shear centre calculated for the C

on (section 1), unsymmetrical I-Section (section 2) and the extended C

section (section 3), section 2 have the smallest value. Hibbeler (2011, 394) clai



producing a smaller value of eccentricity, e and a nearer distance towards the centroid

Lab 1: p15

ar flow direction in an extended C- section flows



mm to the left hand


mm to the left hand side of the section which also lies on the axis of symmetry.

mental and theoretical shear centre found, the percentage error

channel section is 0.96 %.

r centre calculated for the C-Channel

section 2) and the extended C-Channel

section (section 3), section 2 have the smallest value. Hibbeler (2011, 394) claimed that the



ce towards the centroid of

Lab 1: p16


Possible sources of error

The percentage of error is calculated above for each of the sections. By comparing

the percentage errors obtained for all three of the sections, the C-Channel section has

biggest percentage error compared to the other two sections. This can be due to the

misalignment of the parallel flanges of the C-Channel section as the flanges are assumed to

be perfectly aligned and when calculating the theoretical shear centre. Therefore, there are

difference in the location of the shear centre obtained through experimental value and the

theoretical value.

Besides that, there are other few factors that might affect the results obtained during

the experiment. Firstly, the dial gauge readings are presented in analog. The presence of

human parallax error might occur while obtaining the readings. Secondly, the depth of the

dial gauge plunger may vary as the axis of the dial gauge plunger is adjustable. This will

cause some inconsistency towards the readings obtained.

Furthermore, the grip of the section on the specimen is not completely fixed as

assumed in the experiment. The sections might have undergone fatigue after several years

of usage. Lastly, during the experiment while moving the load to different notches, the

load might have oscillated slightly. All these factors can cause the inaccuracy of readings

obtained for the experiment.

Some precautions in future experiments are suggested. Firstly, the apparatus used

should be calibrated from time to time in order to obtain more accurate results. The dial

gauge plunger can be changed to digital instead of analog. This will improve the accuracy

of readings obtained. Other than that, the experiment can be repeated for a few times to

achieve an average result for all sections. Lastly, the loadings should be applied carefully

to decrease the oscillation to the minimum.

Lab 1: p17


5.0 Conclusion

In conclusion, twisting or torsion of the section will not occur if the force applied is

located at the shear centre. Shear centre can be determined by using the shear centre

formula or through plotting a graph of deflection against load position using the

experimental data obtained from the experiment. Furthermore, the percentage error

obtained for C-Channel section, unsymmetrical I-Section and extended C-Channel section

are 24.10 %, 0.38 % and 0.96 % respectively.

Lab 1: p18


6.0 References Hibbeler, R.C. 2011. Mechanics of Materials: Eighth Edition In SI units. Singapore:

Prentice Hall.

P.P.Benham, 2013. Civil Engineering Terms : Easy and Understandable Terms Related to

Civil Engineering. Accessed October 21,

http://www.civilengineeringterms.com/mechanics-of-solids-2/shear-center/


Appendix

Appendix A: Derivation of shear center equation for C

This can be represented by:

Therefore,

The resisting force within any part of the section equals the average shear flow in that

section part multiplied by the area of that section part. The total resultant force (from all

parts) will be equal and opposite to the appl

bcd = ef@2

= eC2 HF@2

= eCHF@


Derivation of shear center equation for C-Channel Section




parts) will be equal and opposite to the applied load. (ie: Sum of the forces equals zero)

V = load applied through the shear centre

S = the shear centre, which is a point through which a

load can be applied such that no rotation of the bean

will occur

= the shear flow set up in the beam to resist the

applied load (the shear flow is a description of how

the resisting force is distributed throughout the

section)

bcd = ef@2

e = W__[506 [4W6

f = Egh?4?012 4i W30W 4i j0V2541 i34? 2H0 V]2251X _[W10

@ = Fkl?4?012 4i W30W 4i 2H0 2 = 2H5Vm10jj 4i 2H0 V]2251X _[W10

The shear flow goes around the corner without any

changes. Hence, bcd is just before D equals

after D which equals bnco (ie: bcd pqr

Where

G[j4 bcd = bnco

W16 bsco = EF bnco #50: bcd ptqu%

vwxy bsco = eCH>@ a RFS

Lab 1: p19



ied load. (ie: Sum of the forces equals zero)

V = load applied through the shear centre

S = the shear centre, which is a point through which a

ch that no rotation of the bean

= the shear flow set up in the beam to resist the

applied load (the shear flow is a description of how

the resisting force is distributed throughout the

j0V2541 30?420

zH4[0 j0V2541

_[W10

The shear flow goes around the corner without any

is just before D equals bco just

pqr)

%

S

Lab 1: p20


{|} = bsco K G30W ~]Cj2< RFS

= eCH>@ K C K 2

= eCIH2>@ a R=S

It is known that when a force is applied to a beam at its shear centre, there is no distortion,

which implies that the moments are balanced. Hence, it follows that ΣM about any point

equals zero.

Take moments about C. (The force due to the shear flow in DB is awkward to determine

but can be disregarded as it acts through C.)

B �� = ;RW125 − V[4Vmz5j0 ?4?012j _4j525\0S

e0 − {|} �HF� − {�� H

F� = ;

e0 = {|} �HF� + {�� H

F�

�4z {|} = {��

~4 e0 = {|}H a R>S

e0 = eCIH2>@ ~]Cj2< R=SW16 R>S

� = ��


Appendix B: Derivation of shear center equation for

Using the same reasoning in Appendix A:


As before:

{�� = bscoR�S K G30W = eC�H

>@ K C K 2

= eCIH2>@ = {��

bncoR�S = ef@2

= eC2 HF@2

= eCHF@

�� bscoR�S = eC�H>@


Derivation of shear center equation for Unsymmetrical I-Section


:

~]Cj2< R:S

a RMS W16 {}� = eCIH2>@ = {��

(Note: The shear flow for FB has not been

shown as we are taking moments about

therefore the shear flow in FB is immaterial.)

a R:S W16 j5?5[W3[8 bscoRIS = e

Lab 1: p21

Section

�� a RNS

(Note: The shear flow for FB has not been

own as we are taking moments about D,

therefore the shear flow in FB is immaterial.)

eCIH>@ a RPS

Lab 1: p22


B �� = ;RW125 − V[4Vmz5j0 ?4?012j _4j525\0S

e0 − {��H + {}�H = ;

e0 = {��H − {}�H e0 = eHI2#C�I − CII%

>@ ~]Cj2< RMSW16 RNS

� = ��#�$� − ��%��


Appendix C: Derivation of shear center equation f



bcoRIS = �e@2� �W2 �H − W

F ��

= eWRH − WSF@

bcoR�S = �ef@2 �}� + �ef

@2 ��

= �eCH2F@2 � + JeWRH −

F@= e

F@ �CH + WRH − WS

bs}� = bcoR�S + bcoRISF

bs}� = bcoRISF

= eWRH − WS>@


Derivation of shear center equation for Extended C-Channel Section



��

a RLS

��

R − WSO

S� a RE;S

� {}� = bs}�C2 = {�� a REE

� {�� = bs��2W = {�� a REF

Note: the shear flow distribution in the leg “a” is

not linear. When “a” is short, the distribution is

almost linear (and the average shear flow is

½ bcoRIS). When “a” approaches h/2 the shear

flow approaches a quadratic distribution (and

the average shear flow is one third of

In this case, as “a” is significantly less than h/2

it is a reasonable approximation to assume that

the average shear flow is ½ bcoRIS

Lab 1: p23

Channel Section

S

EFS

Note: the shear flow distribution in the leg “a” is

not linear. When “a” is short, the distribution is

almost linear (and the average shear flow is

). When “a” approaches h/2 the shear

flow approaches a quadratic distribution (and

the average shear flow is one third of bcoRIS).

In this case, as “a” is significantly less than h/2

it is a reasonable approximation to assume that

R S.

Lab 1: p24


B �� = ;RW125 − V[4Vmz5j0 ?4?012j _4j525\0S

e0 − {}�H − F{��C = ;

e0 = {}�H + F{��C

e0 = eRCIHI2 + FWCHI2 − FWDC2S>@ ~]Cj2< RLS7 RE;S7 REES W16 REFS

� = �� +�� + �� − ��-�� .

Lab 1: p25


Appendix D: Laboratory sheets

sm lab report 1 senior

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