Formula Sheets: Force Method of Analysis | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


GA TE CE 2026 F orm ula Sheet: Structural Analysis
1. Statically Determinate Structures
• Degree of static determinacy:
D
s
=3m+r-3j ( for plane trusses)
D
s
=6m+r-6j ( for space trusses)
where:
– D
s
: Degree of s tatic determinacy
– m : Num b er of mem b ers
– r : Num b er of reactions
– j : Num b er of join ts
• Equilibrium equations (2D):
?
F
x
=0,
?
F
y
=0,
?
M =0
• Equilibrium equations (3D):
?
F
x
=0,
?
F
y
=0,
?
F
z
=0,
?
M
x
=0,
?
M
y
=0,
?
M
z
=0
2. F orce Metho ds (Statically Indeterminate Structures)
• Flexibilit y metho d (force metho d):
d
ij
=
?
M
i
M
j
EI
dx
where:
– d
ij
: Flexibilit y co e?icien t (deflection due to unit forc e)
– M
i
, M
j
: Momen t functions due to applied and redundan t forces
– E : Mo dulus of elasticit y (N/mm²)
– I : Momen t of inertia (mm?)
• Compatibilit y equation:
?
i
+
?
d
ij
F
j
=0
where:
– ?
i
: Displacemen t due to applied loads
– F
j
: Redundan t forces
3. Ener gy Metho ds
• Strain energy:
U =
?
M
2
2EI
dx ( for b ending)
U =
?
N
2
2EA
dx ( for axial)
where:
– U : Strain energy (N·mm)
– M : Bending momen t (N·mm)
– N : Axial force (N)
– A : Cross-sectional area (mm²)
1
Page 2


GA TE CE 2026 F orm ula Sheet: Structural Analysis
1. Statically Determinate Structures
• Degree of static determinacy:
D
s
=3m+r-3j ( for plane trusses)
D
s
=6m+r-6j ( for space trusses)
where:
– D
s
: Degree of s tatic determinacy
– m : Num b er of mem b ers
– r : Num b er of reactions
– j : Num b er of join ts
• Equilibrium equations (2D):
?
F
x
=0,
?
F
y
=0,
?
M =0
• Equilibrium equations (3D):
?
F
x
=0,
?
F
y
=0,
?
F
z
=0,
?
M
x
=0,
?
M
y
=0,
?
M
z
=0
2. F orce Metho ds (Statically Indeterminate Structures)
• Flexibilit y metho d (force metho d):
d
ij
=
?
M
i
M
j
EI
dx
where:
– d
ij
: Flexibilit y co e?icien t (deflection due to unit forc e)
– M
i
, M
j
: Momen t functions due to applied and redundan t forces
– E : Mo dulus of elasticit y (N/mm²)
– I : Momen t of inertia (mm?)
• Compatibilit y equation:
?
i
+
?
d
ij
F
j
=0
where:
– ?
i
: Displacemen t due to applied loads
– F
j
: Redundan t forces
3. Ener gy Metho ds
• Strain energy:
U =
?
M
2
2EI
dx ( for b ending)
U =
?
N
2
2EA
dx ( for axial)
where:
– U : Strain energy (N·mm)
– M : Bending momen t (N·mm)
– N : Axial force (N)
– A : Cross-sectional area (mm²)
1
• Castigliano’s theorem (deflection):
d
i
=
?U
?P
i
where:
– d
i
: Deflection at p oin t of load P
i
– P
i
: Applied load (N)
• Virtual w ork principle:
dW =
?
M ·m
EI
dx =0
where:
– m : Virtual momen t due to unit load
4. Metho d of Sup erp osition
• T otal deflection or reaction:
d
total
=d
1
+d
2
+···+d
n
where:
– d
i
: Deflection due to individual load case i
• T otal momen t or force:
M
total
=M
1
+M
2
+···+M
n
5. Analysis of T russes
• Metho d of join ts (force in mem b er):
?
F
x
=0,
?
F
y
=0 ( at eac h join t)
• Metho d of sections (force in sp ecific mem b er):
F
i
=
M
cut
d
where:
– F
i
: F orce in mem b er i (kN)
– M
cut
: Momen t ab out cut p oin t (kN·m)
– d : P erp endicular distance from c ut p oin t to mem b er (m)
6. Analysis of Arc hes
• Horizon tal thrust for three-hinged arc h:
H =
?
My ·y
EI
dx
?
y
2
EI
dx
where:
– H : Horizon tal thrust (kN)
– M
y
: Momen t due to applied loads (kN·m)
– y : V ertical distance from arc h axis (m)
• Bending momen t at an y section:
M =M
y
-H ·y
2
Page 3


GA TE CE 2026 F orm ula Sheet: Structural Analysis
1. Statically Determinate Structures
• Degree of static determinacy:
D
s
=3m+r-3j ( for plane trusses)
D
s
=6m+r-6j ( for space trusses)
where:
– D
s
: Degree of s tatic determinacy
– m : Num b er of mem b ers
– r : Num b er of reactions
– j : Num b er of join ts
• Equilibrium equations (2D):
?
F
x
=0,
?
F
y
=0,
?
M =0
• Equilibrium equations (3D):
?
F
x
=0,
?
F
y
=0,
?
F
z
=0,
?
M
x
=0,
?
M
y
=0,
?
M
z
=0
2. F orce Metho ds (Statically Indeterminate Structures)
• Flexibilit y metho d (force metho d):
d
ij
=
?
M
i
M
j
EI
dx
where:
– d
ij
: Flexibilit y co e?icien t (deflection due to unit forc e)
– M
i
, M
j
: Momen t functions due to applied and redundan t forces
– E : Mo dulus of elasticit y (N/mm²)
– I : Momen t of inertia (mm?)
• Compatibilit y equation:
?
i
+
?
d
ij
F
j
=0
where:
– ?
i
: Displacemen t due to applied loads
– F
j
: Redundan t forces
3. Ener gy Metho ds
• Strain energy:
U =
?
M
2
2EI
dx ( for b ending)
U =
?
N
2
2EA
dx ( for axial)
where:
– U : Strain energy (N·mm)
– M : Bending momen t (N·mm)
– N : Axial force (N)
– A : Cross-sectional area (mm²)
1
• Castigliano’s theorem (deflection):
d
i
=
?U
?P
i
where:
– d
i
: Deflection at p oin t of load P
i
– P
i
: Applied load (N)
• Virtual w ork principle:
dW =
?
M ·m
EI
dx =0
where:
– m : Virtual momen t due to unit load
4. Metho d of Sup erp osition
• T otal deflection or reaction:
d
total
=d
1
+d
2
+···+d
n
where:
– d
i
: Deflection due to individual load case i
• T otal momen t or force:
M
total
=M
1
+M
2
+···+M
n
5. Analysis of T russes
• Metho d of join ts (force in mem b er):
?
F
x
=0,
?
F
y
=0 ( at eac h join t)
• Metho d of sections (force in sp ecific mem b er):
F
i
=
M
cut
d
where:
– F
i
: F orce in mem b er i (kN)
– M
cut
: Momen t ab out cut p oin t (kN·m)
– d : P erp endicular distance from c ut p oin t to mem b er (m)
6. Analysis of Arc hes
• Horizon tal thrust for three-hinged arc h:
H =
?
My ·y
EI
dx
?
y
2
EI
dx
where:
– H : Horizon tal thrust (kN)
– M
y
: Momen t due to applied loads (kN·m)
– y : V ertical distance from arc h axis (m)
• Bending momen t at an y section:
M =M
y
-H ·y
2
7. Analysis of Beams
• Bending momen t (simply supp orted b eam, UDL):
M
max
=
wL
2
8
where:
– M
max
: Maxim um b ending momen t (kN·m)
– w : Uniformly distributed l oad (kN/m)
– L : Span length (m)
• Maxim um deflection (simply supp orted b eam, UDL):
d
max
=
5wL
4
384EI
• Shear force (p oin t load at midspan):
V =
P
2
where:
– V : Shear force (kN)
– P : P oin t load (kN)
8. Analysis of Cables
• Maxim um tension in cable (parab olic shap e, UDL):
T
max
=
v
H
2
+
(
wL
2
)
2
where:
– T
max
: Maxim um tension (kN)
– H : Horizon tal comp onen t of tension (kN)
– w : Uniform load p er u nit length (kN/m)
– L : Span length (m)
• Cable sag:
f =
wL
2
8H
where:
– f : Sag at midspan (m)
9. Analysis of F rames
• Momen t distribution metho d (balancing momen t):
M
ij
= DF
ij
·M
fixed
where:
– M
ij
: Distributed momen t at join t (kN·m)
– DF
ij
: Distribution fac tor, DF
ij
=
kij
?
k
– k
ij
: Stiffness of mem b er ( k =
4EI
L
for fixed ends)
– M
fixed
: Fixed-end momen t
3
Page 4


GA TE CE 2026 F orm ula Sheet: Structural Analysis
1. Statically Determinate Structures
• Degree of static determinacy:
D
s
=3m+r-3j ( for plane trusses)
D
s
=6m+r-6j ( for space trusses)
where:
– D
s
: Degree of s tatic determinacy
– m : Num b er of mem b ers
– r : Num b er of reactions
– j : Num b er of join ts
• Equilibrium equations (2D):
?
F
x
=0,
?
F
y
=0,
?
M =0
• Equilibrium equations (3D):
?
F
x
=0,
?
F
y
=0,
?
F
z
=0,
?
M
x
=0,
?
M
y
=0,
?
M
z
=0
2. F orce Metho ds (Statically Indeterminate Structures)
• Flexibilit y metho d (force metho d):
d
ij
=
?
M
i
M
j
EI
dx
where:
– d
ij
: Flexibilit y co e?icien t (deflection due to unit forc e)
– M
i
, M
j
: Momen t functions due to applied and redundan t forces
– E : Mo dulus of elasticit y (N/mm²)
– I : Momen t of inertia (mm?)
• Compatibilit y equation:
?
i
+
?
d
ij
F
j
=0
where:
– ?
i
: Displacemen t due to applied loads
– F
j
: Redundan t forces
3. Ener gy Metho ds
• Strain energy:
U =
?
M
2
2EI
dx ( for b ending)
U =
?
N
2
2EA
dx ( for axial)
where:
– U : Strain energy (N·mm)
– M : Bending momen t (N·mm)
– N : Axial force (N)
– A : Cross-sectional area (mm²)
1
• Castigliano’s theorem (deflection):
d
i
=
?U
?P
i
where:
– d
i
: Deflection at p oin t of load P
i
– P
i
: Applied load (N)
• Virtual w ork principle:
dW =
?
M ·m
EI
dx =0
where:
– m : Virtual momen t due to unit load
4. Metho d of Sup erp osition
• T otal deflection or reaction:
d
total
=d
1
+d
2
+···+d
n
where:
– d
i
: Deflection due to individual load case i
• T otal momen t or force:
M
total
=M
1
+M
2
+···+M
n
5. Analysis of T russes
• Metho d of join ts (force in mem b er):
?
F
x
=0,
?
F
y
=0 ( at eac h join t)
• Metho d of sections (force in sp ecific mem b er):
F
i
=
M
cut
d
where:
– F
i
: F orce in mem b er i (kN)
– M
cut
: Momen t ab out cut p oin t (kN·m)
– d : P erp endicular distance from c ut p oin t to mem b er (m)
6. Analysis of Arc hes
• Horizon tal thrust for three-hinged arc h:
H =
?
My ·y
EI
dx
?
y
2
EI
dx
where:
– H : Horizon tal thrust (kN)
– M
y
: Momen t due to applied loads (kN·m)
– y : V ertical distance from arc h axis (m)
• Bending momen t at an y section:
M =M
y
-H ·y
2
7. Analysis of Beams
• Bending momen t (simply supp orted b eam, UDL):
M
max
=
wL
2
8
where:
– M
max
: Maxim um b ending momen t (kN·m)
– w : Uniformly distributed l oad (kN/m)
– L : Span length (m)
• Maxim um deflection (simply supp orted b eam, UDL):
d
max
=
5wL
4
384EI
• Shear force (p oin t load at midspan):
V =
P
2
where:
– V : Shear force (kN)
– P : P oin t load (kN)
8. Analysis of Cables
• Maxim um tension in cable (parab olic shap e, UDL):
T
max
=
v
H
2
+
(
wL
2
)
2
where:
– T
max
: Maxim um tension (kN)
– H : Horizon tal comp onen t of tension (kN)
– w : Uniform load p er u nit length (kN/m)
– L : Span length (m)
• Cable sag:
f =
wL
2
8H
where:
– f : Sag at midspan (m)
9. Analysis of F rames
• Momen t distribution metho d (balancing momen t):
M
ij
= DF
ij
·M
fixed
where:
– M
ij
: Distributed momen t at join t (kN·m)
– DF
ij
: Distribution fac tor, DF
ij
=
kij
?
k
– k
ij
: Stiffness of mem b er ( k =
4EI
L
for fixed ends)
– M
fixed
: Fixed-end momen t
3
• Carry-o v er factor:
CO =0.5 ( for fixed ends )
• Slop e deflection equation:
M
ij
=
2EI
L
(
2?
i
+?
j
-
3?
L
)
+M
fixed
where:
– ?
i
, ?
j
: Rotations at ends i and j (radians)
– ? : Relativ e displacemen t (mm)
4