HKUST Spring 2026
Lecture 09
Statics
[CIVL 2110]
Dr. Xiao PAN
Assistant Professor
Department of Civil and Environmental Engineering
Hong Kong University of Science and Technology
Statics [CIVL 2110]
Lecture 09
equilibrium of a rigid body in 2D
Note:
number of unknowns (reaction forces)
number of equations: (equilibrium conditions
e.g. ΣFx =0, ΣFy =0, ΣΜA =0 )
different cases:
NOT enough!
the structure
should be
stable !
• improper supports (mechanism) → <3 unknowns
• determinate structure
→ 3 unknowns
• indeterminate structure
→ >3 unknowns
(unknown reactions) –
(equilibrium equations) = 0
determinate structure in 2D (proper definition)
3 reaction forces satisfying the following conditions:
1. the 3 reaction forces do NOT cross through the same point
2. the 3 reaction forces are not parallel (this point is at infinite)
2
Statics [CIVL 2110]
Lecture 09
equilibrium of a rigid body in 2D
example 1. simply supported beam
stable
A
B
A
B
B
A
(3)
unstable
A
B
C
3
Statics [CIVL 2110]
Lecture 09
overview
equilibrium of a rigid body in 2D
stability of a rigid body in 2D
basic stable & determinate structures
4
Statics [CIVL 2110]
Lecture 09
equilibrium of a rigid body in 2D
center of
instantaneous
rotation
Note:
the rotation w.r.t. the center of instantaneous
rotation between configurations ABCD and
A'B'C'D' is unique (i.e. θ)
θ
θ
D
C
A
B
5
Statics [CIVL 2110]
Lecture 09
equilibrium of a rigid body in 2D
O
D
CIR: center of
instantaneous
rotation
C
P
A
definition of stable structure
there exists no center of rotation
infinitesimal motion every point P moves in the
direction perpendicular to the position vector OP
B
6
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 2
CIR= center
of rotation
Note:
kinematical approach
stable structure
7
Statics [CIVL 2110]
Lecture 09
overview
equilibrium of a rigid body in 2D
stability of a rigid body in 2D
basic stable & determinate structures
8
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 3. three-hinged arch
internal hinge
two connected rigid bodies
hinged at their ends
infinitesimal motion about the two
center of rotations is allowed! →
infinitesimally this is an unstable
structure later we study this
example from a viewpoint of
equilibrium of forces?
(I)
(II)
(I)
CIR – (I)
CIR – (II)
(II)
9
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 3. three-hinged arch
CIR – (II)
(II)
(I)
(I)
CIR – (I)
improper supports
10
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
internal hinge
200kN
example 3. three-hinged arch
200kN
(II)
B
(I)
(II)
3m
(I)
2m
Ax
Ay
A
4m
2m
4m
Cx
4m
4m
Cy
By
C
Bx
(II)
(1)  Fx = 0  Ax + Cx = 0  ....  Ax = 66.7 kN
( 2 )  Fy = 0  Ay + C y − 200 = 0  Ay = 150 kN
( II )
4
M
(
)
 B =0
( 3)  M A = 0  C y  8 − 2  200 = 0  C y = 50 kN
(3)
C y  4 + Cx  3 = 0 
Cx = −66.7 kN
Cy
Cx
11
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 3. three-hinged arch
(I)
A
200kN
2m
4m
(II)
4m
B
(I)
Ax
200kN
2m
(II)
Cx
B
Ay
Cy
C
(1)  Fx = 0  Ax + Cx = 0
( 2 )  Fy = 0  Ay + C y − 200 = 0
( 3)  M A = 0  C y  8 − 2  200 = 0  C y = 50 kN
Bx
By
(II)
4m
Cx
Cy
( 4 )  M B( II ) = 0 
( 3)

C y  4 = 0  C y = 0 kN  50 kN !
12
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 3. three-hinged arch
2m
200kN
2m
A
B
4m
4m
(II)
δ
Ax
(II)
(I)
δ
(I)
B
Ay
4m
4m
C
Cy
By
Bx
(1)  Fx = 0  Ax + Cx = 0
( 2 )  Fy = 0  Ay + C y − 200 = 0
( 3)  M A = 0  C y  8 − 2  200 = 0  C y = 50 kN
Cx
(II)
Cy
δ
( 4 )  M B ( II ) = 0 
Cx
C y  4 + Cx   = 0 
4
Cx = − C y

if  → 0  Cx = −
13
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 4
5m
5m
C
B
2m
Bx
2m
100kN
100kN
2m
A
2m
D
Ay
Dy
(1)  Fx = 0  Bx + 100 = 0  Bx = −100 kN
( 2 )  Fy = 0  Ay + Dy = 0  Ay = +40 kN
( 3)  M B = 0  2 100 + 5  Dy = 0  Dy = −40 kN
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Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 5 5 m
5m
C
B
Cx
2m
Bx
2m
100kN
100kN
2m
A
2m
D
(1)  Fx = 0  Bx + Cx + 100 = 0  Bx + Cx = −100 kN
( 2 )  Fy = 0  Ay = 0
( 3)  M B = 0  2 100 = 0 ??? Cannot be balanced
Ay
15
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 6. internal hinge
3
internal hinge
4
2m
2m
hinge
A
Ax
E
B
A
x (<2) F
y
Mz
Fx
3m
roller
1m
roller
4(unknown reactions) –
3(equilibrium equations) = 1 ≠0
Fy
Fx
Mx
Ay
C
E
2-x
2m
By
3m
B
C
1m
Cy
Fx, Fy and Mz are called
internal forces: our aim is to
learn to calculate them at
every point
16
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 6. internal hinge
3
• determinate
4
E
B
A
2m
hinge
F = 0
F = 0
M = 0
M = 0
2m
3m
roller
C
1m
roller
x
y
A

E
C
Ey
Cy
additional information provided by
the internal hinge: ΣΜE(Right)= 0
not for the whole structure but for
the right or the left section ONLY
Note: a section cut at an arbitrary
point x yields no useful additional
equation
Ex
But, try this! This will allow the easy calculation of Cy
17
Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 7. cables and pulleys
0.5 m
0.4 m
0.4 m
T
T
0.4 m
60.g
T
T
Ax
Ay
Ey
60.g
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Statics [CIVL 2110]
Lecture 09
basic stable & determinate structures
example 7. cables and pulleys
T
T
0.5 m
T

Ax
T

T = 60  g kN

0.4 m
Ay
because of the pulley
at this point: T = 60  g kN
0.4 m
0.4 m
Ey
60.g
(1)  Fx = 0  Ax − T cos  − T cos  = 0 
Ax = T ( cos  + cos  ) = 911kN
( 2 )  Fy = 0  Ay + E y + T sin  + T sin  − 60 g = 0 
Ay + E y + 60 g ( sin  + sin  ) − 1 = 0 
tan  = 0.5 /1.2   = 22.60
tan  = 0.5 / 0.4   = 51.30
( 3)
Ay + E y = 129kN  Ay = −184.5 kN
( 3)  M A = 0  E y  0.8 + T sin   0.4 + T sin   1.2 − 60 g  1.2 = 0 
E y  0.8 + T ( sin   0.4 + sin   1.2 − 1.2 ) = 0 
E y = 313.5 kN
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Statics [CIVL 2110]
Lecture 09
recap of Lecture 9
A
B
basic stable & determinate
structures:
4
B
A
2m
hinge
2m
3m
roller
E
C
1m
roller
•
simply supported beam
•
internal hinge
•
three-hinged arch
•
cables and pulleys
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