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! Trusses! Vertical Loads on Building Frames! Lateral Loads on Building Frames: Portal

Method! Lateral Loads on Building Frames: Cantilever

Method Problems

APPROXIMATE ANALYSIS OF STATICALLYINDETERMINATE STRUCTURES

2

P1 P2

(a)

Trusses

R1 R2

a

a(b)R1

a

a

V = R1

F1

F2

Fb

Fa

Method 1 : If the diagonals are intentionally designed to be long and slender,it is reasonable to assume that the panel shear is resisted entirely by the tension diagonal,whereas the compressive diagonal is assumed to be a zero-force member.

Method 2 : If the diagonal members are intended to be constructed from largerolled sections such as angles or channels, we will assume that the tension and compressiondiagonals each carry half the panel shear.

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Example 1

Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are to be designed to support both tensile and compressiveforces, and therefore each is assumed to carry half the panel shear. The supportreactions have been computed.

10 kN 20 kN4 m 4 m

3 m

A BC

DEF

4

+ ΣMA = 0: FFE(3) - 8.33cos36.87o(3) = 0

FFE = 6.67 kN (C)

ΣFy = 0:+ 20 - 10 - 2Fsin(36.87o) = 0

F = 8.33 kN

FFB = 8.33 kN (T)

FAE = 8.33 kN (C)

+ ΣMF = 0: FAB(3) - 8.33cos36.87o(3) = 0

FAB = 6.67 kN (T)

ΣFy = 0:+ FAF - 10 - 8.33sin(36.87o) = 0

FAF = 15 kN (T)

10 kN 20 kN4 m 4 m

3 m

A BC

DEF

10 kN

20 kN

V = 10 kN

FFE

FAB

FAE= FFFB= F

θ = 36.87o

10 kN

3 m

A

F20 kN

θ

θ6.67 kN

8.33FAF

10 kN

θA

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C

D

10 kN

3 mθ

θ

ΣFy = 0:+ 10 - 2Fsin(36.87o) = 0

F = 8.33 kN

FDB = 8.33 kN (T)

FEC = 8.33 kN (C)

+ ΣMC = 0: FED(3) - 8.33cos36.87o(3) = 0

FED = 6.67 kN (C)

+ ΣMD = 0: FBC(3) - 8.33cos36.87o(3) = 0

FBC = 6.67 kN (T)

θ = 36.87o

V = 10 kN

θ = 36.87oD

θ6.67 kN

8.33 kNFDC

ΣFy = 0:+ FDC - 8.33sin(36.87o) = 0

FDC = 5 kN (C)

ΣFy = 0:+

FEB = 2(8.33sin36.87o) = 10 kN (T)

Eθ θ

6.67 kN6.67 kN

8.33 kN 8.33 kN

θ = 36.87o

FEB

FED

FBC

F = FEC

F = FDB

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Example 2

Determine (approximately) the forces in the members of the truss shown inFigure. The diagonals are slender and therefore will not support a compressiveforce. The support reactions have been computed.

40 kN 40 kN

4 m

A B C

GHJ

D E

FI

10 kN 20 kN 20 kN 20 kN 10 kN

4 m 4 m 4 m 4 m

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FBH = 0

FIH

FBC

FIC

40 kN

4 m

A

J

10 kN

45o

V = 10 kN

FBH = 0

ΣFy = 0:+ 40 - 10 - 20 - FICcos 45o = 0

FIC = 14.14 kN (T)

+ ΣMA = 0: FJI(4) - 42.43sin 45o(4) = 0

FJI = 30 kN (C)

+ ΣMJ = 0: FAB(4) = 0

FAB = 0

0

0FJA

40 kN

θA

ΣFy = 0:+ FJA = 40 kN (C)

FAI = 0

FJI

FAB

FJB V = 30 kN

ΣFy = 0:+ 40 - 10 - FJBcos 45o = 0

FJB = 42.43 kN (T)

FAI = 0

40 kN

4 m

A B

J I

10 kN 20 kN

4 m

45o

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+ ΣMB = 0:

FIH(4) - 14.14sin 45o(4) + 10(4) - 40(4) = 0

FJH = 40 kN (C)

FBC = 30 kN (T)

+ ΣMI = 0: FBC(4) - 40(4) + 10(4) = 0

FBH = 0

FIH

FBC

FIC V = 10 kN

40 kN

4 m

A B

J I

10 kN 20 kN

4 m

45oB

30 kN0

042.3 kNFBI

45o

45o45o

ΣFy = 0:+ FBI = 42.3 sin 45o = 30 kN (T)

ΣFy = 0:+

FBI = 2(14.1442.3 sin 45o) = 20 kN (C)

C30 kN30 kN

14.14 kN14.14 kNFCH

45o45o

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Vertical Loads on Building Frames

typical building frame

10

w

A B

L(b)

0.21L0.21Lpoint of zero moment

approximate case

w

L

(d)

assumed points of zero moment0.1L 0.1L

model

w

(e)

� Assumptions for Approximate Analysis

0.1L 0.1L0.8L

w

A B

L

Simply supported(c)

column column

girder

w

A BL

(a)

Point ofzeromoment

Point ofzeromoment

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Example 3

Determine (approximately) the moment at the joints E and C caused by membersEF and CD of the building bent in the figure.

B

D

F

A

C

E

6 m

1 kN/m

1 kN/m

12

1 kN/m

1 kN/m

4.8 m

4.8 kN

0.6 m

0.6 kN 2.4 kN

3 kN

0.6(0.3) + 2.4(0.6) = 1.62 kN�m

0.6 m

0.6 kN2.4 kN

3 kN

1.62 kN�m

0.1L=0.6 m 0.6 m

4.8 m

2.4 kN 2.4 kN