Civil Engineering Reference
In-Depth Information
Table 2.1
Summary of basic equilibrium equations
Basic equations
At yield
BENDING
α
r
=
0
◦
M
y
=
A
s
f
y
(
jd
)
M
=
A
s
f
s
(
jd
)
ELEMENT SHEAR
(
q
=
shear flow)
q
y
=
√
n
y
n
ty
q
=
n
cot
α
r
n
y
n
ty
q
=
n
t
tan
α
r
tan
α
r
=
q
=
(
σ
d
h
)sin
α
r
cos
α
r
and
N
=
n
d
v
)
BEAM SHEAR
(
V
=
qd
v
(
N
y
/
d
v
)
n
ty
N
cot
α
r
V
=
V
y
=
d
v
N
y
/
d
v
n
ty
V
=
n
t
d
v
tan
α
r
tan
α
r
=
V
=
(
σ
d
h
)
d
v
sin
α
r
cos
α
r
=
q
(2
A
o
)and
N
=
n
p
o
)
TORSION
(
T
T
y
=
(2
A
o
)
(
N
y
/
p
o
)
n
ty
N
p
o
(2
A
o
)cot
α
r
T
=
(
N
y
/
p
o
)
n
ty
T
=
n
t
(2
A
o
)tan
α
r
tan
α
r
=
T
=
(
σ
d
h
)(2
A
o
)sin
α
r
cos
α
r
for the three cases of element shear, beam shear and torsion. Comparison of the three sets of
five equations for element shear, beam shear and torsion shows that they are basically the same.
The five equations for beam shear are simply the five equations for element shear multiplied by
a length d
v
. The five equations for torsion are simply those for element shear multiplied by the
area 2
A
o
. Hence it is only necessary to understand the geometric and algebraic relationships
of the set of five equations for element shear.
Within each set of five equations for element shear, beam shear, and torsion the three in the
left-hand column are derived directly from the three equilibrium conditions. The geometric
