Torsion in Girders

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25 Νοε 2013 (πριν από 4 χρόνια και 5 μήνες)

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Torsion in Girders

A2

A3

M
u

=
w
u
l
n
2
/24

M
u

=
w
u
l
n
2
/10

M
u

=
w
u
l
n
2
/11

B2

B3

The

beams

framing

into

girder

A
2
-
A
3

transfer

a

moment

of

w
u
l
n
2
/
24

into

the

girder
.

This

moment

acts

the

longitudinal

axis

of

the

girder

as

torsion
.

A

torque

will

also

be

induced

in

girder

B
2
-
B
3

due

to

the

difference

between

the

end

moments

in

the

beams

framing

into

the

girder
.

Girder A2
-
A3

A2

A3

Torsion Diagram

A2

A3

Strength of Concrete in Torsion

cp
cp
c
c
u
P
A
f
T
T
2
4
3
'

As for shear, ACI 318 allows flexural members be designed for
the torque at a distance ‘d’ from the face of a deeper support.

where,

A
cp

= smaller of b
w
h + k
1
h
2
f

or b
w
h + k
2
h
f
(h
-

h
f
)

P
cp

= smaller of 2h + 2(b
w
+ k
1
h
f
) or 2(h + b
w
) + 2k
2
(h
-

h
f
)

k
1

k
2

Slab one side of web

4

1

Slab both sides of web

8

2

Redistribution of Torque

p. 44 notes
-
underlined paragraph

When

redistribution

of

forces

and

moments

can

occur

in

a

statically

indeterminate

structure,

the

maximum

torque

for

which

a

member

must

be

designed

is

4

times

the

T
u

for

which

the

torque

could

have

been

ignored
.

When

redistribution

cannot

occur,

the

full

factored

torque

must

be

used

for

design
.

In

other

words,

the

design

T
max

=

4
T
c

if

other

members

are

available

for

redistribution

of

forces
.

Torsion Reinforcement

Stresses

induced

by

torque

are

resisted

with

closed

stirrups

and

longitudinal

reinforcement

along

the

sides

of

the

beam

web
.

The

distribution

of

torque

along

a

beam

is

usually

the

same

as

the

shear

distribution

resulting

in

more

closely

spaced

stirrups
.

Design of Stirrup Reinforcement

u
y
t
oh
T
f
A
A
t
s
28
.
1

"
12
8
'
3
)
2
(
4

h
c
w
y
t
v
P
f
b
f
A
A
s
A
oh

= area enclosed by the centerline of the closed
transverse torsional reinforcement

A
t

= cross sectional area of one leg of the closed ties
used as torsional reinforcement

s
t

= spacing required for torsional reinforcment only

s
v

= spacing required for shear reinforcement only

S

= stirrup spacing

t
v
s
s
s
1
1
1

Design of Longitudinal
Reinforcement

t
t
h
y
cp
c
t
h
t
l
s
A
P
f
A
f
s
P
A
A
'
5
y
w
t
t
f
b
s
A
where
25
,

max. bar spacing = 12”

minimum bar diameter = s
t
/
24

A
l

= total cross sectional area of the
required to resist torsion

P
h

= perimeter of A
oh

Cross Section Check

To prevent compression failure due
to combined shear and torsion:

c
A
P
T
d
b
V
f
oh
h
u
w
u
'
5
.
7
2
7
.
1
2
2

Design of Torsion Reinforcement

for Previous Example
p. 21 notes

Design

the

torsion

reinforcement

for

girder

A
2
-
A
3
.

30 ft

30 ft

30 ft

30 ft

24 ft

24 ft

24 ft