video conference exercises 2014

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TST2601VIDEO CONFERENCE EXERCISES 2014

1

VIDEO CONFERENCE EXERCISES

SECTIONAL PROPERTIES, SHEAR FORCE AND BENDING MOMENT

1. The figure below shows a built up section. Calculate the following with XX and YY as

reference point.

a. The position of the centroid of the section.

b. The second moment of area about horizontal and vertical axis (Ixx and Iyy).

c. The radius of gyration, rx and ry.

d. The elastic section modulus.

e. The plastic section modulus.

SOLUTION

(a)




2

44

2323

44

2323

2

2

32.10033.100320444806333.28583325930810000

)7.1945(700701012/1)57.19(12001012012/1

32.27833.278320484286333.58334945081440000

)57.39(700107012/1)7.3960(12001201012/1

)(

7.191900

)45(700)5(1200

7.391900

)5(700)60(1200

19007001200

1070)12010(

cmmm

x x x x I

cmmm

x x x x I

Ae I I

b

mm x

mm y

mm

x x A

YY

XX

GG XX

(c)

mm A

I r

mm A

I r

YY y

XX x

78.2291900

100320433

03.1211900

27832043

(d)

341006.7017.39

278320433mm x

y

I Z XX

ex

(e)

362250

45125112516000

5.47)9510(5.7)1510(20)1080(

1510

150

80095010

108002

1900

)(10)10(802/1

mm Z

Z

x x x Zp

mm P

P

P

P A

p

p




3

2. The figure below shows a built-up section. Calculate the following with XX and YY as

reference point.

(a) The position of the centroid of the section

(b) The second moment of area about horizontal axis (Ixx)

(c) The radius of gyration, r x

(d) The elastic section modulus.

SOLUTION

(a)

marksmmmark x

marksmmmark y

marksmm

marks x x x A

236.4012.4743

33.338.125633.334000502000

221.3112.4743

67.468.125667.464000102000

22.47438.125640002000

24

4080100

2

120100

2

2

Y

Y

X

m

40mm

100mm

100mm

X20mm

Ф40mm




4

(b)

4

24

23

23

2

859.29169.29185938.3003891256804.95604622.14222222.89972867.66666

]21.3167.468.125664

40[21.3167.464000

36

801001021.312000

12

20100

cmmm

I

Ae I I

xx

GG xx

(c)

mm A

I r xx x 81.24

2.4743

69.2918593

(d)

37.9351421.31

69.2918593mm

y

I Z xx

ex

3. A non-standard I section is shown in the Figure below.

a) Decide the point of centroid

b) Decide the second moment of area about x-x(horizontal going through the centroid), xx I

c) Decide the second moment of area about y-y(vertical going through the centroid), yy I

d)

Decide radius of gyration, y x r and r

e) Decide the elastic section modulus

f) Decide the plastic section modulus

70mm

20mm

20mm

10mm

110mm

80mm




5

SOLUTION

(a)

)1()(91.504400

2200048000154000

)2

11(4400

102200608001101400

)1()(55)1(440022008001400

2011010802070

2

mark bottom frommm y

marks x x x

y

mark axis yabout l symmetricamm xmark mm

x x x A

(b)

marksmm

marks I

Ae I I

xx

GG xx

231.9183030

82.368198133.7333348.6610267.42666634.488827967.46666

2141091.502200122011091.5060800

12801091.501101400

122070

4

2

3

2

3

2

3

2

(c)

)2

11(67.2796666

33.221833367.666667.571666

)2

11(12

)110(20

12

)10(80

12

)70(20

4

333

marksmm

marks I yy

(d)

)1(21.254400

67.2796666

)1(68.454400

31.9183030

mark mm A

I r

mark mm A

I r

YY y

XX x

(e)

)1(73.18037791.50

31.9183030 3

ma x

mark mm y

I Z XX

ex




6

(f)

)1(1800001260003200022000

)2(9020704010801020110)(.

int,20modint:

3 mark mm

marks x x x x x x x Zparea

totatheof half is pothisat areathebottom frommm yat isulus plasticof Po Note

4. Draw the shear force and bending moment diagrams for the beams shown below. Determine

the values and positions of maximum shear force and bending moment. Determine the position

of contraflexure.

SOLUTION

kN R

R R

F

kN R

R

R

R

M

D

D A

y

A

A

A

A

D

33.10867.81190

13555

0

67.81

7359

04053309

0)3(135)6(55)9(

0

3m 3m 3m

135kN55kN

B

A D

C




7

kN S

kN S

kN S

kN S

kN S

kN S

FORCES SHEAR

D

right C

left C

right B

left B

A

33.108

33.1081355567.81

67.265567.81

67.265567.81)

67.81

67.81

)(

)(

(

)(

0

02.325)3(55)6(67.81

01.245)3(67.81

0

D

C

B

A

BM

kNm BM

kNm BM

BM

MOMENT BENDING

3m 3m 3m

135kN55kN

B

A D

C

81.67

26.67

108.33

245.01

325.02




8

Maximum shear force = 108.33 kN at D.

Maximum bending moment = 325.02 kNm at C

There is no point of contraflexure.

5. Draw the shear force and bending moment diagrams for the beam shown below. Determine

the values and positions of maximum shear force and bending moment. Determine position/s of

contraflexure.

SOLUTION

kN R

R R

F

kN R

R

R R

M

D

D A

y

A

A

A

A

D

11.24189.203445

445180265

0

89.203

18359

0245159090245)6(265)9(

0




9

kN right S

kN left S kN S

kN right S

kN left S

kN S

FORCES SHEAR

D

D

C

B

B

A

9011.241)3(3026589.203)(

11.151)3(3026589.203)(11.6126589.203

11.6126589.203)(

89.203)(

89.203

0

99.1342

)3(30245)6(265)9(89.203

34.183245)3(265)6(89.203

34.428)3(265)6(89.203

67.611)3(89.203

0

2

E

D

C

B

A

BM

kNm BM

kNmright BMC

kNmleft BM

kNm BM

BM

MOMENT BENDING

POINT OF CONTRAFLEXURE (POC)

POC occurs at range 0 ≤ x ≤ 9 between C and D

Equation of moment beween C and D =

m xor m x

xor x

x

x

a

acbbuse

x POC at

x

x x x x

x x x

009.808333.0

30

39.12189.118

30

39.12189.118

30

39.12189.118

30

83.1473489.118

)15(2

)10)(15(4)89.118(89.118

2

4

01089.11815

1089.11815

)3612(1524579526589.203

2

)6(30245)3(26589.203

2

2

2

2

2

2




10

POC = 8.01m from A, it is within the range 0 ≤ x ≤ 9.

SFD

BMD

203.89

90

151.11

61.11

134.99

183.34

428.34

611.67




11

6. A cantilever beam 2 m long has a rectangular cross-section 50 mm wide and 150 mm deep. It

carries a uniformly distributed load of 5kN/m over the whole span and in addition to this an

axially applied compressive force of 3kN.

(a) Calculate the maximum resultant direct stress in the beam, and

(b) for this section, plot a stress distribution diagram

(c) Also calculate the position of the neutral axis.

SOLUTION

(a) Maximum bending moment due to uniformlydistributed load = -5 x 2 x 1 = 10 kNm

Ixx of the cross section is46

33

1006.1412

15.005.0

12m x

xbd

y = 0.075mm (symmetrical section)

Ixx

y M maxmax

MPa x

x

x33.53075.0

1006.14

10106

3

max

Since the cross section is symmetrical about the neutral axis,the magnitude of the stress at the top

and bottom of the section at the wall will be the same.

Consider the axially applied load of 3 kN

)(4.015.005.0

3000C MPa

x

A

P




12

(b)

(c)

Using similar triangle to get the position of neutral axis:

mm x

x

x x

x x

x x

56.7566.106

5.8059

5.805966.106

93.5273.535.8059

93.52)150(73.53

150

93.5273.53

Neutral axis is at 75.56mm from top

±=

0.4MPa C

0.4MPa C

53.33MPa T

53.33MPa T 52.93MPa T

53.73MPa C

video conference exercises 2014

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