Analysis of Statically Determinate
Structures
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Idealized Structure
Principle of Superposition
Equations of Equilibrium
Determinacy and Stability
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Beams
Frames
Gable Frames
Application of the Equations of Equilibrium
Analysis of Simple Diaphragm and Shear
Wall Systems Problems
1
Classification of Structures
• Support Connections
weld
stiffeners
weld
typical “pin-supported”
connection (metal)
typical “roller-supported”
connection (concrete)
typical “fixed-supported”
connection (metal)
typical “fixed-supported”
connection (concrete)
2
pin support
pin-connected joint
fixed-connected joint
torsional spring support
fixed support
torsional spring joint
P
P
A
B
L/2
actual beam
L/2
B
A
L/2
L/2
idealized beam
3
Table 2-1 Supports for Coplanar Structures
Type of
Connection
(1)
(2)
θ
Idealized
Symbol
Light
cable
θ
Reaction
Number of Unknowns
One unknown. The reaction is a
force that acts in the direction of
the cable or link.
θ
One unknown. The reaction is a
force that acts perpendicular to
the surface at the point of contact.
rollers
F
rockers
(3)
F
(4)
F
One unknown. The reaction is a
force that acts perpendicular to
the surface at the point of contact.
One unknown. The reaction is a
force that acts perpendicular to
the surface at the point of contact.
4
Type of
Connection
Idealized
Symbol
Reaction
(5)
Fy
Fx
Number of Unknowns
Two unknowns. The reactions
are two force components.
Smooth pin or hinge
(6)
slider
F
M
Two unknowns. The reactions
are a force and moment.
fixed-connected collar
(7)
Fx
fixed support
M
Fy
Three unknowns. The reactions
are the moment and the two force
components.
5
• Idealized Structure.
3m
B
3m
B
F
4m
F
A
actual structure
4m
A
idealized structure
6
C
B
D
joist
slab
girder
A
column
idealized framing plan
fixed-connected beam
idealize beam
fixed-connected overhanging beam
Idealized beam
7
idealized framing plan
idealized framing plan
8
• Tributary Loadings.
stringer
veihicle
slab
slab
girder
girder
stringer
floor beam
floor beam
deck girder
pier
9
spandrel
beam
beam
2nd floor
joist slab
joist
beam
supported slab
foundation
wall
1st floor
stairs
landing
slab on grade
wall
footing
column
basement
spread
footing
10
One-Way System.
4m
B
A
A
1m
0.5 kN/m2
C
C
B
E
D
1m
D
4m
F
2m
2m
idealized beam
2 kN
1 kN
2 kN
4m
F
idealized framing plan
D
2 kN
1m
E
1 kN/m
C
1m
1 kN
B
F
2m
2m
idealized girder
11
column
girder
concrete slab is
reinforced in two
directions, poured
on plane forms
L1/2
A
beam
L2
B
A
C
D
L1/2
L1
L1
E
F
Idealized framing plan
for one-way slab action
requires L2 / L1 ≥ 2
12
L2/L1 = 1
Two-Way System.
4m
B
A
A
0.5
4m
45o
kN/m2
2m
C
1 kN/m
45o
A
4m
B
B
2m
4m
idealized beam, all
D
C
D
2m
idealized framing plan
L2/L1 = 1.0 < 2
6m
B
A
2m
45o
1kN/m
1 kN/m
45o
4m
D
C
idealized framing plan
A
B
2m
2m 2m
A
C
2m
2m
idealized beam
13
Principle of Superposition
P = P1 + P 2
Two requirements must be imposed for the principle
of superposition to apply :
=
d
P1
1. The material must behave in a linear-elastic
manner, so that Hooke’s law is valid, and therefore
the load will be proportional to displacement.
σ = P/A
δ = PL/AE
d
+
d
P2
2. The geometry of the structure must not
undergo significant change when the loads are
applied, i.e., small displacement theory applies.
Large displacements will significantly change
and orientation of the loads. An example would
be a cantilevered thin rod subjected to a force at
its end.
14
Equations of Equilibrium
ΣFx = 0
ΣFy = 0
ΣFz = 0
ΣMx = 0
ΣMy = 0
ΣMz = 0
N
M
M
V
N
V
internal loadings
15
Determinacy and Stability
• Determinacy
r = 3n, statically determinate
r > 3n, statically indeterminate
n = the total parts of structure members.
r = the total number of unknown reactive force and moment components
16
Example 2-1
Classify each of the beams shown below as statically determinate or statically
indeterminate. If statically indeterminate, report the number of degrees of
indeterminacy. The beams are subjected to external loadings that are assumed to
be known and can act anywhere on the beams.
hinge
hinge
17
SOLUTION
r = 3, n = 1, 3 = 3(1)
Statically determinate
r = 5, n = 1, 5 - 3(1) = 2 Statically indeterminate to the second degree
hinge
r = 6, n = 2, 6 = 3(2)
r = 10, n = 3, 10 - 3(3) = 1
Statically determinate
Statically indeterminate to the first degree
18
Example 2-2
Classify each of the pin-connected structures shown in figure below as statically
determinate or statically indeterminate. If statically are subjected to arbitrary
external loadings that are assumed to be known and can act anywhere on the
structures.
19
SOLUTION
r = 7, n = 2, 7 - 3(2) = 1
r = 9, n = 3, 9 = 3(3)
Statically indeterminate to the first
degree
Statically determinate
20
r = 10, n = 2, 10 - 6 = 4
degree
r = 9, n = 3, 9 = 3(3)
Statically indeterminate to the fourth
Statically determinate
21
Example 2-3
Classify each of the frames shown in figure below as statically determinate or
statically indeterminate. If statically indeterminate, report the number of degrees
of indeterminacy. The frames are subjected to external loadings that are assumed
to be known and can act anywhere on the frames.
B
C
A
D
22
SOLUTION
B
C
A
D
r = 9, n = 2, 9 - 6 = 3
Statically indeterminate to the third degree
r = 15, n = 3, 15 - 9 = 6
Statically indeterminate to the sixth degree
23
• Stability
r < 3n, unstable
r > 3n, unstable if member reactions are concurrent
or parallel or some of the components form
a collapsible mechanism
Partial Constrains
P
A
P
A
MA
FA
24
Improper Constraints
O
O
B
A
C
d
P
B
P
C
C
d
FA
P
A
B
A
A
FA
FC
FB
B
FB
P
C
FC
25
Example 2-4
Classify each of the structures in the figure below as stable or unstable. The
structures are subjected to arbitrary external loads that are assumed to be known.
B
A
A
B
hinge
A
C
B
B
A
A
B
C
C
D
26
SOLUTION
B
A
The member is stable since the reactions are non-concurrent and nonparallel.
It is also statically determinate.
hinge
A
C
B
The compound beam is stable. It is also indeterminate to the second degree.
A
B
C
The compound beam is unstable since the three reactions are all parallel.
27
A
B
The member is unstable since the three reactions are concurrent at B.
B
A
C
D
The structure is unstable since r = 7, n = 3, so that, r < 3n, 7 < 9. Also, this can
be seen by inspection, since AB can move horizontally without restraint.
28
Application of the Equations of Equilibrium
D
B
A
P1
Dx
P1
Ex
C
By
Ay
P1
P2
Ax
Dy
By
E
P2
Dx
Bx
Ay
Bx
Ey
Ax
Ex
P2
Cx
Cx
r = 9, n = 3, 9 = 3(3);
statically determinate
29
P1
P1
A
B
Ay
Bx
Ax
By
P2
C
P1
B
Bx
Ay
Ax
P2
Cx
P2
Cy
Cx
Cy
r = 6, n = 2, 6 = 3(2);
statically determinate
30
Example 2-5
Determine the reactions on the beam shown.
A
0.3 m
3m
150 kN
60o
1m
B
2m
70 kN•m
31
SOLUTION
265 kN
60o
0.3 m
A
3m
1m
70 kN•m
B
2m
265 sin 60o = 229.5 kN
Ax
Ay
+ ΣF = 0:
x
+ ΣMA = 0:
+
265 cos 60o = 132.5 kN
0.3 m
ΣFy = 0:
3m
1 m By
70 kN•m
Ax - 132.5 = 0: Ax = 132.5 kN , →
By(4) - (229.5)(3) + (132.5)(0.3) -70 = 0
By = 179.69 kN, ↑
Ay - 229.5 + 179.69 = 0
Ay = 49.81 kN , ↑
32
Example 2-6
Determine the reactions on the beam shown.
15 kN/m
5 kN/m
A
12 m
33
SOLUTION
(1/2)(12)(10) = 60 kN
15 kN/m
5 kN/m
A
12 m
10 kN/m
5 kN/m
Ax
MA
Ay
(5)(12) = 60 kN
12 m
4m
6m
+ ΣF = 0:
x
+
ΣFy = 0:
+ ΣMA = 0:
Ax = 0
Ay - 60 - 60 = 0
Ay = 120 kN , ↑
MA - (60)(4) - (60)(6) = 0
MA = 600 kN•m
34
Example 2-7
Determine the reactions on the beam shown. Assume A is a pin and the support at
B is a roller (smooth surface).
B
7 kN/m
3m
A
4m
2m
35
B
SOLUTION
7 kN/m
3m
A
4m
28 kN
2m
3m
Ax
Ay
90o-56.3o = 33.7o
NB
tan-1(3/2) = 56.3o
2m
6m
+ ΣMA = 0:
-28(2) + NBsin 33.7(6) + NBcos 33.7(3) = 0
NB = 9.61 kN
+ ΣF = 0:
x
Ax - NBcos 33.7 = 0; Ax = 9.61cos 33.7 = 8 kN , →
+
ΣFy = 0:
Ay - 28 + 9.61cos33.7 = 0
Ay = 22.67 kN , ↑
36
Example 2-8
The compound beam in figure below is fixed at A. Determine the reactions at A,
B, and C. Assume that the connection at pin and C is a rooler.
6 kN/m
hinge
A
C
B
6m
8 kN•m
4m
37
SOLUTION
6 kN/m
hinge
8 kN•m
A
C
B
6m
4m
36 kN
Ax
8 kN•m
Bx Bx
MA
Ay
By
3m
By
Cy
6m
Member BC
Member AB
+ ΣMB = 0:
Cy(4) - 8 = 0
Cy = 2 kN , ↑
+ ΣMA = 0:
MA - 36(3) + 2(6) = 0
MA = 96 kN•m
+ ΣF = 0:
x
Bx = 0
+ ΣF = 0:
x
Ax - B = 0 ; Ax = Bx = 0
ΣFy = 0:
Ay - 36 + 2 = 0
Ay = 34 kN , ↑
+
ΣFy = 0:
Cy - By = 0;
By = Cy = 2 kN , ↑
+
38
Example 2-9
The side girder shown in the photo supports the boat and deck. An idealized
model of this girder is shown in the figure below, where it can be assumed A is a
roller and B is a pin. Using a local code the anticipated deck loading transmitted
to the girder is 6 kN/m. Wind exerts a resultant horizontal force of 4 kN as
shown, and the mass of the boat that is supported by the girder is 23 Mg. The
boat’s mass center is at G. Determine the reactions at the supports.
1.6 m
1.8 m
2m
6 kN/m
4 kN
0.3 m
C
D
G
A
roller
B
pin
39
SOLUTION
1.6 m
1.8 m
2m
6 kN/m
4 kN
0.3 m
C
D
G
A
B
roller
pin
+ ΣF = 0:
x
4 - Bx = 0
Bx = 4 kN , ←
+ ΣMB = 0:
6(3.8) = 22.8 kN
22.8(1.9) -Ay(2) + 225.6(5.4)
-4(0.3) = 0
Ay = 630.2 kN , ↑
1.9 m
4 kN
0.3 m
C
D
G
Bx
2m
Ay
23(9.81) kN = 225.6 kN
5.4 m
+
By
ΣFy = 0:
-225.6 + 630.2 - 22.8 + By = 0
By = 382 kN , ↑
40
Example 2-10
Determine the horizontal and vertical components of reaction at the pins A, B,
and C of the two-member frame shown in the figure below.
8 kN
5
4
3
3 kN/m
B
C
2m
2m
1.5 m
A
2m
41
SOLUTION
Member BC
8 kN
3 kN/m
5
4
B
3
+ ΣMC = 0:
-By(2) +6(1) = 0
By = 3 kN , ↑
Member AB
C
2m
2m
1.5 m
+ ΣMA = 0:
-8(2) - 3(2) +Bx(1.5) = 0
Bx = 14.7 kN , ←
+ ΣF = 0:
A
6 kN
2m
x
Bx
8 kN
1m
(4/5)8 By
Cx
1m
By
Cy
Bx
(3/5)8
1.5 m
Ay - (4/5)8 - 3 = 0
Ay = 9.4 kN , ↑
Member BC
+ ΣF = 0:
x
Ax
Ay
2m
+
Ax + (3/5)8 - 14.7 = 0
Ax = 9.87 kN , →
ΣFy = 0:
+
ΣFy = 0:
Cx - Bx = 0; Cx = Bx = 14.7 kN , ←
3 - 6 + Cx = 0 ; Cy = 3 kN , ↑
42
Example 2-11-1
From the figure below, determine the horizontal and vertical components of
reaction at the pin connections A, B, and C of the supporting gable arch.
B
3m
15 kN
3m
C
A
3m
3m
43
SOLUTION
B
3m
15 kN
3m
Ax
C
A
Ay
3m
3m
Cx
Cy
Entire Frame
+ ΣMA = 0:
C y (6) − 15(3) = 0
Cy = 7.5 kN , ↑
+
ΣFy = 0:
Ay + 7.5 = 0
Ay = -7.5 kN , ↓
44
B
3m
15 kN
Bx 3.75 kN= Bx
By
B
3m
7.5 kN = By
3m
3m
Ax
C
A
7.5 kN
3m
3m
Member AB
Member BC
+ ΣMB = 0: 15(3) + Ax (6) + 7.5(3) = 0
+ ΣF = 0:
x
+ ΣF = 0:
x
Ax = -11.25 kN , ←
Cx
7.5 kN
3.75 − C x = 0
Cx = 3.75 kN
− 11.25 + 15 − Bx = 0
Bx = 3.75 kN , ←
+
ΣFy = 0:
− 7.5 + By = 0
By = 7.5 kN
45
Example 2-11-2
The side of the building in the figure below is subjected to a wind loading that
creates a uniform normal pressure of 1.5 kPa on the windward side and a suction
pressure of 0.5 kPa on the leeward side. Determine the horizontal and vertical
components of reaction at the pin connections A, B, and C of the supporting gable
arch.
2m
2m
B
3m
3m
C
4m
A
4m
3m
3m
wind
46
SOLUTION
2m
2m
B
3m
3m
C
4m
A
4m
3m
3m
wind
A uniform distributed load on the
windward side is
(1.5 kN/m2)(4 m) = 6 kN/m
A uniform distributed load on the
leeward side is
(0.5 kN/m2)(4 m) = 2 kN/m
B
6 kN/m
2 kN/m
6 kN/m
C
A
3m
2 kN/m
3m
3m
3m
47
25.46 sin 45 8.49 sin 45
25.46 kN
B
8.49 kN
25.46 cos 45
45o
45o
8.49 cos 45
3m
18 kN
6 kN
1.5m
Ax
Ay
1.5
3m
Cx
1.5 Cy
Entire Frame
+ ΣMA = 0:
+
ΣFy = 0:
-(18+6)(1.5) - (25.46+8.49)cos 45o(4.5) - (25.46 sin 45o)(1.5)
+ (8.49 sin 45o)(4.5) + Cy(6) = 0
Cy = 24.0 kN , ↑
Ay - 25.46 sin 45o + 8.49 sin 45o 3 + 24 = 0
Ay = -12.0 kN
48
25.46 sin 45
25.46 kN
25.46 cos 45
By
18 kN
Ax
Bx
45o
8.49 sin 45
8.49 kN
Bx
1.5
45o
By
3
8.49 cos 45
6 kN
1.5 1.5
1.5
Cx
Ay= 12.0 kN
Cy = 24.0 kN
Member AB
+ ΣMB = 0: (25.46 sin 45o)(1.5) + (25.46cos 45o)(1.5) + (18)(4.5) + Ax(6) + 12(3) = 0
Ax = -28.5 kN
+ ΣF = 0: -28.5 + 18 + 25.46 cos 45o - B = 0
x
+
ΣFy = 0:
Member CB
+ ΣF = 0:
x
x
Bx = 7.5 kN , ←
-12 - 25.46sin 45o + By = 0
By = 30.0 kN , ↑
7.5 + 8.49 cos 45o + 6 - Cx = 0
Cx = 19.50 kN , ←
49
Analysis of Simple Diaphragm and shear Wall Systems
A
B
B
A
A
B
B A
F
F/8
agm
r
h
p
F/8
dia
f
o
o
F/8
r
F/8
F/2
F/8
A
A
F/8
F/8
F/8
F/8
F/8
A
A
F/8
F/8
F/8
F/8
F/8
floor diaphragm
F/8
F/2
50
roof diaphragm
A
C
D
C A
D B
Wind F
F/16
B
second floor
diaphragm
F/16
roof
F/4
shear walls
agm
r
h
p
dia
F/16
F/16
A
F/16
F/16
F/16
F/16
3F/16
F/2
3F/16
2 st
floor
3F/16 F/16
F/16
3F/16
3F/16
B
3F/16
3F/16
3F/16
F/4
1 st floor
3F/16
F/4
F/4
3F/16
F/4
F/4
51
Example 2-12
Assume the wind loading acting on one side of a two-story building is as shown
in the figure below. If shear walls are located at each of the corners as shown and
flanked by columns, determine the shear in each panel located between the floors
and the shear along the columns.
20 m
30 m
A
1.2 kPa
C
D
B
C A
D B
3m
4m
4m
3m
0.8 kPa
52
SOLUTION
20 m
30 m
A
1.2 kPa
C
D
C A
D B
3m
FR2
FR1
B
3m
12 kN
4m
4m
12 kN
roof
FR2 /2
= 48 kN
12 kN
A
FR1 /2 = 32
FR2 = 1.2(103) N/m2 (20 m)(4 m) = 96 kN
FR2 /2 = 48
12 kN
F/8 = 12 kN
12 kN
32 kN 12 kN
12 kN
s
t
2 floor
32 kN
12 kN
32 kN
0.8 kPa
FR1 = 0.8(103) N/m2 (20 m)(4 m) = 64 kN
agm
r
h
p
dia
32 + 48 kN
32 kN
32 kN
B
32 kN
32 kN
32 kN
1 st floor
40 kN
32 kN
FR1 /2
= 32 kN
40 kN 32 kN
40 kN
40 kN
53
12 kN
Fv
Fv
4m
+ ΣM = 0:
Fv(3) - 12(4) = 0
Fv = 16 kN
12 kN
3m
32 kN
F´v
F´v
32 kN
+ ΣM = 0:
4m
F´v(3) - 32(4) = 0
F´v = 42.7 kN
3m
54