Torsion in Girders

A2

A3

Mu = wuln2/24 Mu = wuln2/10 Mu = wuln2/11

B2

B3

The beams framing into girder A2-A3 transfer a moment of wuln2/24 into the girder. This moment acts about the longitudinal axis of the girder as torsion.

A torque will also be induced in girder B2-B3 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

cpccu P

AfTT

2

43 '

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,Acp = smaller of bwh + k1h2

f or bwh + k2hf(h - hf)Pcp = smaller of 2h + 2(bw+ k1hf) or 2(h + bw) + 2k2(h - hf)

k1 k2

Slab one side of web 4 1Slab both sides of web 8 2

Redistribution of Torque

p. 44 notes-underlined paragraphWhen 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 Tu 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 Tmax = 4Tc 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

ytoh

T

fAA

ts28.1 "128'3

)2(4 h

cw

ytv P

fb

fAAs

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

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

st = spacing required for torsional reinforcment only

sv = spacing required for shear reinforcement only

S = stirrup spacing

tv sss111

Design of Longitudinal Reinforcement

t

th

y

cpc

t

htl s

AP

f

Af

s

PAA

'5

y

w

t

t

f

b

s

Awhere

25,

max. bar spacing = 12”minimum bar diameter = st/24

Al = total cross sectional area of the additional longitudinal reinforcement required to resist torsion

Ph = perimeter of Aoh

Cross Section Check

To prevent compression failure due to combined shear and torsion:

cA

PTdb

V foh

hu

w

u '5.72

7.1

2

2

Design of Torsion Reinforcementfor Previous Example p. 21 notes

Design the torsion reinforcement for girder A2-A3.

30 ft30 ft30 ft30 ft

24 ft

24 ft

24 ft