C5303

1 ANALYSIS OF STATICALLY 2D FRAME STRUCTURE

1.1 Introduction

A frame structure is a load bearing unit constructed by joining members with pin/hinge or welding. A main characteristic of such a structure is that it is built with a series of basic triangles (? = ).

Other characteristics are it must fulfill the equilibrium and stability requirements.

1.2 Stability

Stable structure = It will not undergo any change in shape when loaded, consisting of a minimum number of members for maintaining its original shape. The basic stable form is a triangle.

Generally a stable structure fulfills the equation: m 2j – 3

j = number of joints

3 = M = 0 Fx = 0 FY = 0 and M = 0

Examples

C C’ D D’ m = 3, j = 3 C m = 4, j = 4

2j - 3 = 2(3) - 3 = 3 2j - 3 = 2(4) - 3 = 5 = m m < 5

A B A B

Stable Uns table

rectangle

Stable Stable Unstable

1.3 Indeterminacy

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Indeterminacy is a structural property with respect to the method of solution based on static equilibrium equations. For example, a statically determinate structure should have a minimum number of support reactions (r = 3) for equilibrium requirement, and can be solved by using three basic equilibrium equations ( M = 0, FX = 0, FY = 0 ).

External Statically determinate

( = ) InternalFrame structure

External Statically indeterminate

( > ) Internal

a) Externally determinate fulfills r = 3 ( r = R, H, M )

b) Internally determinate fulfills m = 2j - 3

(a) + (b) m + r = 2j

c) Externally indeterminate fulfills r > 3 ( r = R, H, M )

d) Internally indeterminate fulfills m > 2j - 3 (c) + (d) m + r > 2j

**A Statically indeterminate structure consists of more members/reactions than the minimum required for equilibrium. However it is generally stronger than statically determinate structures and more difficult to solve. Examples

a) b)

r = 3 m = 11 j = 7 r = 4 m = 11 j = 7 m + r = 2j (m + r = 15) > (2j = 14)

Stat. determinate Externally stat. indeterminate

c) d)

r = 3 m = 14 j = 8 (m + r = 17) > (2j = 16) ?????

Internally stat. indeterminate

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SUMMARY: CONCEPT OF FRAME STRUCTURE

M + r < 2j

Internal m = 2j - 3 D = 0

Stablem 2j -

3

Unstablem < 2j - 3

r < 3

StaticallyIndeterminate

Internal m > 2j - 3 Di > 0

External r > 3 DO > 0

Statically determinate

Internal m + r > 2j D > 0 & External

Frame Structure

External r = 3 D = 0

Internal m + r = 2j D = 0 & External

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1.4 Analysis of Internal Forces – Statically Determinate Structure

Analysis to determine all external forces (reactions) and internal forces (member forces).

Solution methods: Equilibrium of joints method

Section method

Graphic method (next article)

Tension Coefficient method (next article)

1.4.1 Equilibrium of Joints/Hinges Method

Pre-requisite:

b) Two type of equilibriums for pin joint

Horizontal : FX = 0 ….. (1)

Vertical: FY = 0 ….. (2)

Max. of 2 unknowns can be solved

b) Components of inclined forces & Trigonometric properties

F F FY FY FX FX

FX FX

FY FY

F F

FX = F cos ** must be wrt. horizontal

FY = F sin

c) Reactions: FX = 0 Fy = 0 M = 0

Analysis steps

i. Determine all support reactions

ii. Start at any joint with 2 unknowns.

ii. Draw all forces acting at the joint and assume tension type ( )

iii. Solve the forces by using FX = 0 dan FY = 0

iv. Repeat for other joints with 2 unknowns.

* If the frame & and loads are symmetrical solve for either side only.

Examples

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Determine all reactions and member forces (magnitude and type) for the structure shown below.1.

100 kN Solution

T/balas: MB = 0: 3RA = 100(1.5) RA = 50 kN

FY = 0: RA + RB = 100 kN RB = 50 kN

@ By symmetry : RA = RB = 50 kN

HA = 600 600

FX = 0: HA = 0 kN

Joint A: FAC FY = 0: 50 + FAC sin 60O = 0 FAC = - 57.74 kN (C)

60O FX = 0: FAB + FAC cos 60O = 0 FAB = 28.87 kN (T) A FAB

50

Joint B:

FBC FY = 0: 50 + FBC sin 60O = 0 FBC = - 57.74 kN (C)

60O B @ By symmetry : FBC = FAC = - 57.74 kN (C)

50

2. a) State all characteristics for the frame below

b) Determine the magnitude and value of member forces by using the method of joint equilibrium

20 kN 40 kN 20 kN

1,5 m

AX

2 m 2 m 2m AY BY

10 kN

Solution

3 mRB

RA

28,87

A B

C D E

FG

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a) Characteristics r = 3 m = 11 j = 7

m = 2j - 3 = 11 stable, int. & ext. statically determinate

b) Member forces

Reactions : MB = 0: 6Ay = 20(5) + 10(4) + 40(3) + 20(1) Ay = 46,67 kN

FY = 0: Ay + By = 90 kN By = 43,33 kN FY = 0: Ax = 0

tan = 1,5/1 = 56,3o

Joint A: FAC FY = 0: FAC sin 56,3o + 46,67= 0 FAC = - 56,1 kN (C)

FX = 0: FAG + FAC cos 56,3o = 0 FAG = 31,1 kN (T) A FAG

46,67

Ttk. C 20

C FCD FY = 0: FCG sin 56,3o + (- 56,1 sin 56,3o) + 20 = 0

FCG = 32,1 kN(T)

- 56,1 FCG FX = 0: FCD + FCG cos 56,3o = - 56,1 cos 56,3o

FCD = - 48,9 kN (C) Joint G

32,1 FDG FY = 0: FDG sin 56,3o + 32,1 sin 56,3o = 10

FDG = - 20,1 kN (C) 31,1 G FFG FX = 0: FFG + FDG cos 56,3o = 32,1 cos 56,3o + 31,1

FFG = 60,1 kN (T) 10

Joint B:

FBE FY = 0: FBE sin 56,3o + 43,33 = 0 FBE = - 52,1 kN (C)

B FBF FX = 0: FBF + FBE cos 56,3o = 0

43,33 FBF = 28,9 kN (T) Joint E

20 FDE E FY = 0: FEF sin 56,3o + 20 + (- 52,1) sin 56,3o = 0

FEF = 28,1 kN (T)

FEF - 52,1 FX = 0: FDE + FEF cos 56,3o = - 52,1cos 56,3o

FDE = - 44,5 kN (C) Joint F

FDF 28,1

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FY = 0: FDF sin 56,3o + 28,1 sin 56,3o = 0

60,1 F 28,9 FDF = - 28,1 kN (C) Check : FX = 0: FDF cos 56,3o + 60,1 = 28,1 cos 56,3o + 28,9

FDF = - 28,1 kN (C) OK

20 kN 40 kN 20 kN

C - 48,9 D - 44,5 E

- 56,1 - 28,1 32,1 -20,1 28,1 - 52,1 (Optional)

A 31,1 G 60,1 F 28,9 B

AY 46,67 BY 43,33 10 kN

By using the method of joint eqilibrium determine the values and magnitudes of forces for all members in the structure below.3.

B C D

3 m

A G F E

20 kN 60 kN

3 @ 4 m = 12 m

Solution

B C D MA = 0: 8RF = 20(4) + 60(12)

RF = 100 kN () 3 m

FY = 0: RA + 20 + 60 = RF

A RA = 20 kN

HA G F E FX = 0: HA = 0 RA RF

20 kN 60 kN tan = 3/4 = 36,870

Member forces – Anggapkan semua daya adalah tegangan

Joint A

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FAB FY = 0: FAB = 20 (T)

A FAG FX = 0: FAG = 0

RA = 20

Joint B FY = 0: FBG sin + 20 = 0 FBG = - 33,33 kN (C)

B FBC FX = 0: FBC + FBG cos = 0 FBC = 26,67 kN (T)

20 FBG

Joint C

C FCD FY = 0: FCG = 0 26,67 FX = 0: FCD = 26,67 (T)

FCG

Joint G FY = 0: FDG sin + (-33,33 sin ) = 20

-33,33 FDG FDG = 66,67 kN (T) G 0 FFG FX = 0: FDG cos + FFG = - 33,33 cos

FFG = - 80 kN (C) 20

Joint F

FDF FY = 0: FDF + 100 = 0 FDF = - 100 kN (C) FX = 0: FEF = - 80 (C) - 80 F FEF

100

Joint E FY = 0: FDE sin = 60 FDE = 100 kN (T)

FDE

E Check: FX = 0 FDE cos + (- 80) = 100 cos - 80 = 0

- 80 OK

60

B 26,67 C 26,67 D

20 0 100

-33,33 66,67 -100 A E

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0 G - 80 F - 80

20 20 100 60

By using the method of joint eqilibrium determine the values and megniudes of forces for all members in the structure below.4.

HB B C

25 kN RB

h

HA 45O 45O A E D 600

3 m 3 m 60 kN

Solution

Reactions: h/3 = tan 450 h = 3 m

MB = 0 : 3HA + 60 cos 600 (3) = 60 sin 600 (6) HA = 73,92 kN ( )

FY = 0 : RB = 60 sin 600 = 51,96 kN ()

FX = 0 : HB + 25 = HA + 60 cos 600 HB = 78,92 kN ()

Member forces

Joint A FY = 0 : FAB = 0 FAB

FX = 0 : FAG + 73,92 = 0 FAG = - 73,92 kN (C) 73,92 FAG

A

Joint B

RB = 51,96

78,92 FBC FY = 0 : FBE sin = 51,96 FBE = 73,48 kN (T) B 45O

FX = 0 : FBC + FBG cos = 78,92 FBC = 26,96 kN (T) FBE

0

Joint E FCG FY = 0 : FCG + 73,48 sin = 0 FCG = - 51,96 kN (C)

73,48 FX = 0 : FFG = 73,48 cos + (- 73,92)

- 73,92 G FFG FFG = - 21,96 kN (C)

Joint D

FCD

D FY = 0: FCD sin = 60 sin 600 FCD = 73,48 kN (T)

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FDE 600 FX = 0: FDE + FCD cos = 60 cos 600

60 FDE = - 21,96 kN (C)

Joint C C

26,96 25 FY = 0: FCE + 73,48 sin = 0 FCE = - 51,96 kN (C) 45O

Check FX = 0 : 73,48 cos - 26,96 - 25 = 0 OK FCE 73,48

B 26,96 C 78,92 25

51,96 73,48 73,48 0 -51,96 73,92

A -73,92 E -21,96 D 600

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1.4.2 Analysis of Internal Forces – Section Method

Concept: A section is made through members (figure below). Then determine member forces from equilibrium of sub-frame ( FY = 0, FX = 0, M = 0 )

Hint:

P1 P2 P3

C D E

h A B

G F

C FCD D C FCD D E

h FDG FDG

A B

a G FFG F G FFG F

Mint.bot = 0 Ftop

Mint.top = 0 Fbot

FY @ FX = 0 Fint

Ftop

Fbot

Fint

P1 P2 P3

RA RB

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RA RB

Equilibrium of left sub-frame:

For MG = 0: FCD.h + RA.a = P1.a/2 FCD = … [Mint.bot = 0]

For MD = 0: FFG.h + P1 a = RA 3a/2 FFG = … [Mint.top = 0]

For FY = 0: FDG sin + RA = P1 FDG = (P1 - RA)/sin

@ FX = 0 .....................

Or, from equilibrium of right sub-frame.

For MG = 0: FCD.h + RB.2a = P2.a/2 + P3.3a/2 FCD = … [Mint.bot = 0]

For MD = 0: FFGh + P3a = RB 3a/2 FFG = … [Mint.top = 0]

For FY = 0: FDG sin + P2 + P3 = RB FDG = (RB - P2 - P3 )/sin

The selection of sub-frame and equilibrium equations strictly individuals.

Analysis steps:

i. Determine all support reactions

ii. Draw a section through members ( 3 unknowns).

iii. Draw all forces acting on the frame and assume tension type ( )( -ve answers means compression )

iii. Solve the forces by using FX = 0 dan FY = 0

iv. Repeat for other joints with 2 unknowns.

* If the frame & and loads are symmetrical solve for either side only.

The method of section seems to be more appropriate when dealing with small frames or when only a few numbers of member forces are required.

Examples

By using the section method determine the magnitudes and types of forces FAB and FAC for the frame below.1.

B D

6 m

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A F C E

6 kN 3 kN

3 @ 8 m = 24 m

Solution

Reactions: MF = 0 24RA = 6(16) + 3(8) RA = 5 kN

FY = 0 RA + RF = 9 kN RF = 4 kN

= tan-1 6/8 = 36,87O

Member forces - Draw a section through AB and AC as shown below and consider the equilibrium of left sub-frame, assume all forces are of tension type.

B FAB FY = 0 FAB sin + 5 = 0 FAB = - 8,33 kN (C)

FX = 0 FAC + FAB cos = 0 FAC = 6,67 kN (T)

A FAC C RA = 5 kN

Note: If the right sub-frame is considered, more works are required.

B D FAB

6 m

A F C E

FAC 6 kN 3 kN RF =4 kN

3 @ 8 m = 24 m

FY = 0: FAB sin + 6 + 3 = 4 FAB = - 8,33 kN (C)

FX = 0: FAC + FAB cos = 0 FAC = 6,67 kN (T) either one

@ MB = 0: FAC(6) + 3(8) = 4(16)FAC = 6,67 kN (T)By using the section method determine the magnitudes and types of forces FBD, FCD and FCE for the frame in example 1. (By considering left sub-frame).2.

Solution – Draw a section through BD, CD and CE as shown

B FBD D FY = 0: FCD sin + 5 = 6 FCD = 1,67 kN (T)

FCD MD = 0: FCE(6) + 6(8) = 5(16) FCE = 5,33 kN (T)

A E MC = 0: FBD(6) + 5 (8) = 0 FBD = - 6,67 kN (C)

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C FCE

RA = 5 kN 6 kN @ FX = 0: FBD + FCD cos + FCE = 0

FBD = - 6,67 kN (C)

3.By using the section method determine the magnitudes and types of forces FBD, FCD and FCE for the frame in example 1 (by considering right sub-frame).

Check it yourself ….!

1.5 Zero-Force Member

In certain circumstances the use of zero force members are necessary to prevent premature buckling of compression members or vibration in tension members.

These members can be identified without having to do any calculation as follows.

Two cases

i) At the junction of perpendicular non-concurrent planar forces – no external force

a) b)

ii) At the junction of non-perpendicular non-concurrent planar forces – no external force

a)

Example Identify all zero force members in the frame below. (No calculation is necessary)

30 kN 60 kN

B C D

3 m

A G F E

0 0

0

00

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20 kN

3 @ 4 m = 12 m

Solution

Zero force members FAG, FDE, DEF, FFG and FCG.

END OF TOPIC 1

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