UNIVERSITI TUN HUSSEIN ONN MALAYSIA

ANSWER SCHEME FOR FINAL EXAMINATION SEMESTER I

SESSION 2008/2009

SUBJECT : STRUCTURAL ANALYSIS

SUBJECT CODE : BFC 3023

COURSE : 3 BFF

EXAMINATION DATE : NOVEMBER 2008

DURATION : 3 HOURS

INSTRUCTION : ANSWER FOUR (4) QUESTIONS FROM SIX (6) QUESTIONS

THIS PAPER CONSISTS OF TWENTY THREE (23) PAGES

BFC 3023

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Q1 (a) A truss is loaded as shown in Figure Q1. Prove that the truss is statically determinate.

m = 9 j = 6 r = 3 m = 2j – 3 9 = 2(6) – 3 9 = 9

Proven truss is statically stable and determinate (2 marks)

(b) By using the method of virtual work with the Modulus of Elasticity, E = 200 Gpa

and the cross sectional area for all members, A = 600 mm2, determine:

(i) reactions at support A and C

ΣMA = 0 RC (8) – 10(8) – 15(4) = 0 RC = 17.5 kN ( ↑ ) ΣFy = 0 RA + 17.5 – 10 – 15 – 10 = 0 RA = 17.5 kN ( ↑ ) ΣFy = 0 HA = 0 (2 marks) (ii) internal force for each member

(3 marks)

10 kN 15 kN 10 kN

A

B C

4 m 4 m

3 m

D E F 0

-10 -12.5

10

-7.5 12.5

-17.5

0

-10

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(iii) vertical displacement at joint B and horizontal displacement at joint F

vertical displacement at joint B

(3 marks)

horizontal displacement at joint B

(3 marks)

A

B C

D E F 1

0 -0.625

-0.5

0.375 -0.625

0.375

0

0.5

0.375 0.375

1

1

A

B C

D E F 0

0 -0.84

0.67

0.5 0.84

-0.5

0

-0.67

0.5 1 0.5

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Member L F µ1 µ2 Fµ1L Fµ2L

(m) (kN)

AB 4 10 0.67 -0.5 26.80 -20.00

BC 4 0 0 0 0.00 0.00

CD 3 -17.5 -0.5 0.375 26.25 -19.69

DE 4 -10 -0.67 0.5 26.80 -20.00

EF 4 0 0 1 0.00 0.00

FA 3 -10 0 0 0.00 0.00

AE 5 -12.5 -0.84 -0.625 52.50 39.06

BE 3 -7.5 0.5 0.375 -11.25 -8.44

BD 5 12.5 0.84 -0.625 52.50 -39.06

173.60 -68.13

(8 marks)

Vertical displacement at joint B = ∆By = )200(600

)1000(60.173 = 1.45 mm ( ↓ )

(2 marks)

Horizontal displacement at joint F = ∆Fx = )200(600

)1000(13.68− = -0.57 mm = 0.57 mm (→)

(2 marks)

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Q3 (a) State four (4) cases that will cause sway on the rigid frame and sketch all of them.

(4 Marks) Solution:

1) Horizontal load

2) Vertical load does not symmetry 3) Unsymmetrical frame system 4) Unequal support

(Each answer ½ Mark) (Each sketch ½ Mark)

(b) Briefly explain the procedures of analysis of rigid sway frame to determine the end

moment of each joint. (5 Marks)

Solution:

In general, solution should do deep 2 level and this stages then combined (overlap principal) to get solution another situation in fact.

(1 Mark)

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First stage, framework with no sway caused by constraint that imposed to top end column. In this regard moment distribution can calculate as usual. Consider, final moment that was found was M1 (moment no sway).

(1 Mark) Second stages sway framework due to Q counter direction to constraint discharged. In this regard problem have burden horizon to extension only. Speak omen final that was found was M2 (moment with sway).

(1 Mark) Then actual moment, M = M1 + x .M2

(1/2 Mark) Where x is correction value if ∆ is considered has something other value than the real value. Hence, if F is the ability those found from the case no sway and Q is from the sway case, then:

(1 Mark) 0. =− QxF

Q

Fx = (1/2 Mark)

(c) Figure Q3 shows the non-sway rigid frame which is pinned at A and D and fixed at C. The modulus of elasticity and second moment of area for each member is as shown.

(i) Calculate the moment at all joints using modified moment distribution

method. (10 Marks)

Figure Q3

3 m 3 m 6 m

6 m

EI EI

EI

50 kN

25 kN/m

A B C

D

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Solution: Distribution Factor, DF

Stiffness AB, KAB = KBA L

EI3= EIEI

5.06

)(3 == (1/2 Mark)

Stiffness AB, KBC = KCB L

EI4= EIEI

667.06

)(4 == (1/2 Mark)

Stiffness AB, KBD = KDB L

EI3= EIEI

5.06

)(3 == (1/2 Mark)

At A,

DFAB 0+

=AB

BA

K

K1

05.0

5.0=

+=

EI

EI (1/2 Mark)

At B,

DFBABDBCBA

BA

KKK

K

++= 30.0

5.0667.05.0

5.0=

++=

EIEIEI

EI (1/2 Mark)

DFBCBDBCBA

BC

KKK

K

++= 40.0

5.0667.05.0

667.0=

++=

EIEIEI

EI (1/2 Mark)

DFBDBDBCBA

BD

KKK

K

++= 30.0

5.0667.05.0

5.0=

++=

EIEIEI

EI (1/2 Mark)

At C,

DFCB ∞+

=CB

BC

K

K0

667.0

667.0=

∞+=

EI

EI (1/2 Mark)

At D,

DFDB 0+

=DB

BD

K

K1

05.0

5.0=

+=

EI

EI (1/2 Mark)

Fixed End Moment, FEM

FEMAB kNmL

Pab225

6

)3)(3(502

2

2

2

−=−=−=

(1/2 Mark)

FEMCB kNmL

bPa225

6

)3()3(502

2

2

2

===

FEMBC kNmWL

7512

)6(25

12

22

−=−=−=

(1/2 Mark)

FEMCB kNmWL

7512

)6(25

12

22

===

FEMBD = FEMDB = 0 (1/2 Mark)

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Moment Distribution

Joint A B C D Member AB BA BC BD CB DB CF 0.5 0.5 0.5 0.5 0 0.5 DF 1 0.30 0.40 0.30 0 1 FEM Dist.

-225 225

225 -45

-75 -60

0

-45

-75 0

0 0

CO Dist.

0 0

112.5 -33.75

0

-45

0

-33.75

-30 0

0 0

CO Dist.

0 0

0 0

0 0

0 0

-22.5

0

0 0

End Moment

0

258.75

-180

-78.75

-127.5

0

(3 Marks) (1 Mark)

3 m 3 m 6 m

6 m

A B C

D

258.75 180

78.75

127.5

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(ii) Calculate the reaction at support A, C and D. (6 Marks)

Solution: Support A ∑ = 0BM ∑ = 0YF

075.258)3(50)6( =+−AR 05013.18 1 =+−− BR

)(13.186

15075.258 ↓−=+−= kNRA )(13.6813.18501 ↑=+= kNRB

(1 Mark) (1 Mark) Support C ∑ = 0CM ∑ = 0YF

05.127)3)(6(25)6(2 =−−BR 0)6(2525.96 =+− cR

)(25.966

5.5772 ↑==BR )(75.5325.96150 ↑=−= kNRc

(1 Mark) (1 Mark) Support D ∑ = 0BM

)(13.136

75.78075.78)6( ←−=−=⇒=−− kNHH DD

(1 Mark) ∑ = 0XF

)(13.13013.13 →=⇒=+− kNHH BB (1 Mark)

258.75 kNm

3 m 3 m

50 kN

RA RB1

3 m 3 m

25 kN/m

RB2 RC

127.5 kNm

6 m

HD

HB

78.75 kNm

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Q4 (a) Dengan menggunakan garis imbas yang telah dibina kita dapat:

(i) Mengetahui kedudukan beban bergerak yang akan memberi kesan terbesar (sama ada momen, daya ricih atau tindakbalas) pada sesuatu keratan rentang.

(ii) Mendapatkan nilai (sama ada momen, daya ricih atau tindak balas) pada

sesuatu keratan berdasarkan ordinat garis imbas.

( 2 marks ) Q4 (b) (i) Assume the beam as below: X 1 A C B 2.5m 7.5m

( 1mark) Buat keratan pada titik C, X 1 SY RA kiri SY kanan RB

Ambil kiri bahagian kiri dari keratan rasuk. Ambil momen dari B bagi mendapatkan tindakbalas di A. B RA x 10 – 1x(10-X) = 0 Dari itu, RA = 1 – X/10 Jumlahkan daya pada bahagian kiri: ∑ RA – 1 – SY = 0 , Dari itu, SY = RA – 1 atau = 1 – X/10 –1 = - X/10 …(1) ( 1 mark) Jika bahagian kanan di ambil kira: (beban satu unit sekarang bergerak di bahagian kanan) ∑ RB – 1 + SY = 0 di mana RB = 1- RA = X/10 SY = 1 – RB = 1 – X/10 ………..(2) ( 1 mark) Dari pengiraan di atas didapati terdapat 2 nilai Sy iaitu: Untuk 0 < X < 2.5 SY = - X/10 ( 0.5 mark) 2.5< X < 10 SY = 1 – X/10 ( 0.5 mark)

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Jika Sy lawan X di plot, 0.75 SY +ve X -ve 0.25 2.5m 7.5m \ ( 1 mark)

Garis imbas untuk daya ricih di C Jika berdasarkan masalah di atas, kita gantikan a= 2.5m dan b= 7.5m dan L=10m, maka kita akan dapati ordiant tertinggi adalah seperti berikut: (1 mark) b/L + ve -ve a/L Q4 (b(ii) Assume beam as below:

( 1 mark)

Potong C kepada 2 bahagian A C B

BFC 3023

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Jika 1 unit berada antara 0 hingga 4 m X 1 Mc RA

B RA x 10 = 1 x (10-X) Dari itu, RA = 1 – X/10 Ambil momen dari C, C Mc + 1 x ( 4-X) – RA x 4 = 0 Mc + 4 – X – 4 + 4X/10 = 0

Dari itu, Mc = 6X/10 = 3X/5 ………(1) ( 1 mark) Jika sekarang 1 unit berada antara 4m dan 10 m, A X 1 Mc RB

RB = 1 –RA = 1 – 1 + X/10 = X/10 (0.5 mark)

Ambil momen dari C, C Mc + 1(X – 4) – RB x 6 = 0 Mc + X – 4 – 6X/10 = 0 atau Mc = 4 – 2X/5 ……………….(2) ( 1 mark) Jika persamaan 1 dan 2 diplotkan, rajah berikut di dapati: 12/5 4 – 2X/5 3X/5 X 0 4m 10m

( 1 mark ) Dari masalah di atas jika rasuk di ganti dengan a =4m, dan b=6m ( 0.5 mark) a b L

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Ordinat tertinggi boleh didapati sebagai Mc = ab/L ab/L

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Q6 Forces in the members assuming the frame is pin-jointed;

2 marks

Beban Euler ;

kNx

xxxP ABE 740

10)24(

10120020062

42

== π 2 marks

kN61610x)24(

10x1000x200xP

62

42

BCE =π= 2 marks

740

P

P

P

ABE

ABAB ==ρ ;

616

P

P

P

BCE

BCBC ==ρ 2 marks

83.0616/

740/ ==P

P

BC

AB

ρρ

OR BCAB ρρ 83.0= 1 mark

6.21211024

1012003

4

===x

X

L

Ik AB

AB ; 17681024

1010003

4

===x

X

L

Ik BC

BC 2 marks

2.1=∴BC

AB

k

k 1 mark

Instability occur when ∑MB = 0 and the load will be a critical load (Pcr). ∑MB = MBA + MBC

= ( ) BBCBCABAB sksk θ+ 2 marks Instability condition;

0=+=∑ BCBCABABB skskM OR 02.1 =+ BCAB ss 2 marks

P

A

B

C

VA = P/2 VC = P/2

P P

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Refer to Table 2 6 marks

Let instability occur when ρBC = 2.24 and ρAB = 1.86 1 mark ∴ PAB = 1.86 x 740 = 1376 kN 1 mark

PBC = 2.24 x 616 = 1380 kN 1 mark

1st Trial 2nd Trial 3rd Trial 4th Trial 5th Trial 6th Trial ρBC 0 2 2.4 2.2 2.28 2.24 ρAB 0 1.66 2 1.83 1.89 1.86 sBC 4 0.14 -1.3 -0.52 -0.82 -0.665 sAB 4 1.08 0.14 0.6 0.47 0.552

1.2sAB 4.8 1.3 0.17 0.72 0.56 0.662 1.2sAB + sBC 8.8 1.44 -1.13 0.2 -0.26 -0.003