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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.
3
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
5
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
6
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
7
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
8
+ 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
9
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
11
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