Torsion

introduction
• A moment acting about a
longitudinal axis of the
member is called a torque,
twisting moment or torsional
moment, T.
• Torsion may arise as the
result of:
(a) Primary or equilibrium
torsion: occurs when the
external load has no
alternative to being resisted
but by torsion. Examples:
curved girders and the three
structures shown in Figure.
 introduction
• secondary or compatibility torsion:
in statically indeterminate
structures from the requirements
of continuity. Neglecting this
torsion will not cause problems
because: (1) the shear and
moment capacities of the beam
are not reduced by small amounts
of torque, and (2) the stressing of
adjacent members as the beam
twists permit a redistribution of
forces to these members and
reduces the torque that must be
supported by the beam.
 Examples of torsion

1. Floor systems: compatibility torque

(perimeter beams supporting one or two way
slab systems).
2. Floor system: equilibrium torque (circular
beams).
3. Circular tie beams in mosques.
Analyze the shown structure
BM slab
BM diagrams
Axial force
Mx
My
Torsion diagram
 Shearing stresses due to torsion in
un-cracked members
• Torsion subject is usually neglected in teaching courses of
strength of materials. The average designer does not worry
about torsion although most structures are subjected to
torsional stresses. However, reducing the factor of safety over
the past years resulted in increasing situations were torsional
failures occur with the result that torsion is a more common
problem. The following is an introduction to the elastic
torsion theory as given in mechanics of materials books.
 Behavior of Circular Sections
• Although circular sections are rarely a consideration in normal concrete
construction, a brief discussion serves as a good introduction to the
torsional behavior of other types of sections.
• The basic assumptions are:
1. Plane sections perpendicular to the axis of a circular member remains
plane after torque is applied.
2. Radii of section stay straight (without warping).
As a result of applying the torsion shearing stresses
are set up on cross
sections perpendicular
to the axis of the bar as
shown in Fig.
 Behavior of Circular Sections
• Shear stress is equal to shear strain times the
shear modulus in the elastic range. If r is the
radius of the element, J = πr4 /2 its polar
moment of inertia, and max is the maximum
elastic shearing stress due to elastic twisting
moment T, then from basic strength
(mechanics) of material courses
T r
 max =
J
 Behavior of rectangular sections
Such sections do not fall under the assumptions stated before. They
warp when a torque is applied and radii don't stay straight. As a
result axial as well as circumferential shearing stresses are
generated. For a rectangular member, the corner elements do not
distort at all (corners=0) and the maximum shear stresses occur at
the midpoints of the long sides as shown in Figure. These
complications plus the fact that reinforced
concrete sections are neither homogeneous nor
isotropic make it difficult to develop exact
mathematical formulations based on the physical
models.
•  
 Behavior of rectangular sections
• From the mathematical theory of elasticity the magnitude of
the maximum shear stress , at midpoint of the long side, due
to a torque T as a function of the ratio y to x (long to short
sides) is given by
T
 max =
x y
2

• Where α varies from 0.208 for y/x=1 (square bar) to 0.333 for
y/x= (infinity wide plate)
 Hollow members
• Consider a thin‑wall tube subjected to a torsion T as shown in
Fig. 8.5. If the thickness of the tube is not constant and varies
along the perimeters of the tube, then equilibrium of an
element like that shown in Figure b requires:
V AB = V CD   1 t 1 dx =  2 t 2 dx   1 t 1 =  2 t 2  q

• Where q is referred to as the shear flow and is constant.

 Hollow members
In order to relate the shear flow q to the torque T, consider an element of
length ds as shown. This element is subjected to a force qds and

but rds = T  P rqdx

= the
twice area of the shaded
triangle, then

q T
T = 2q A o   = 
t 2A t
where A is the area enclosed by theo middle of the wall of the tube. From
o
the above equation max occurs where t is the least.
 Examples
• Read example 7.1 in textbook
• Example 2: compute the shear stress, , at the wall and at the
lower flange in the section shown below, due to an applied
torque of 1000kN.m.
Principal stresses due
to torsion
Principal tensile stresses eventually cause
cracking that spirals around the body, as
Shown by the line A-B-C-D-E
In reinforced concrete such a crack would
Cause failure unless it was crossed by
reinforcement. This generally takes the form
of longitudinal bars in the corners and closed
stirrups.
 Principal stresses due to
torsion and shear
The two shear stresses components add on one
side face (front side) and counteract each other
on the other. As the result inclined cracking
starts on AB and extends across the flexural
tensile face. If bending moments are large, the
cracks will extend almost vertically across the
back face. The flexural compression zone near
the bottom prevents the cracks from extending
full height.
Behavior of RC members subjected to torsion
When a concrete member is loaded in
pure torsion, shear stresses develop.
One or more cracks (inclined) develop
when the maximum principal tensile
stress reaches the tensile strength
of concrete. The onset of cracking
causes failure of an unreinforced
Member. Furthermore the addition of longitudinal steel without
stirrups has little effect on the strength of a beam loaded in pure
torsion because it is effective only in resisting the longitudinal
component of the diagonal tension forces.
A rectangular beam with longitudinal bars in the corners and closed
stirrups can resist increased load after cracking as shown in figure.
Behavior of RC members subjected to torsion

After the cracking of a reinforced beam, failure may occur in

several ways. The stirrups, or longitudinal reinforcement, or
both, may yield, or, for beams that are over-reinforced in
torsion, the concrete between the inclined cracks may be
crushed by the principal compression stresses prior to yield of
the steel. The more ductile behavior results when both
reinforcements yield prior to crushing of the
concrete.
Figure shows that ultimate strength of
rc beams were the same for solid and hollow
beams having the same reinforcement.
 Combined torsion,
moment and shear
Test results for beams without stirrups
loaded with various ratios of torsion
and shear are plotted in figure. The
lower envelope of data is given as:

Tc 2 V c 2
( ) ( ) =1
T cu V cu
Where Vcu =inclined cracking shear in the absence of torque.
Tcu = the cracking torque in the absence of shear.
 Space Truss Analogy
Theory
Assumptions:
1. Both solid and hollow members are
considered as tubes.
2. After cracking the tube is idealized
as a hollow truss consisting of
closed stirrups, longitudinal bars in
the corners, and compression
diagonals approximately centered
on the stirrups. The diagonals are
idealized as being between the
cracks that are at angle , generally
taken as 45 degrees for RC.
 q T
= 
The cracking pure torsion t 2Ao t
Knowing that the principal tensile stress equal to the shear stress for
elements subjected to pure shear, thus the concrete will crack
when the shear stress equal to the tensile capacity of cross section.
If we use conservatively 0.333√fc as tensile strength of concrete in
biaxial tension-compression, and remembering that Ao must be
some fraction of the area enclosed by the outside perimeter of the
full concrete cross section Acp. Also, the value of t can, in general,
be approximated as a fraction of the ratio Acp/Pcp, where Pcp is the
perimeter of the cross section. Then, assuming a value of Ao
approximately equal to 2 Acp /3, and a value of t=3 Acp/4Pcp. Using
these values in Eq. above yields:
f c A 2cp
T cr =
Note: tensile strength under biaxial 3 pcp
compression And tension is less than in uniaxial tension (≈0.67).
 Tc 2 V c 2
The cracking torsion ( ) ( ) =1
T cu V cu
In combined shear and torsion, if T=0.25Tcr , the reduction in the inclined cracking
shear is:
Vc 0.25T cr 2
( ) = 1 ( )  0.97
This was deemed to be negligible. V cu T cr
The threshold torsion below which torsion can be ignored in a solid cross section is:

For isolated beam, Acp is thearea A 2cp by the perimeter of the section
f cenclosed
including the area T = holes. Pcp is the perimeter of the section. For a beam
ofcrany
12 pcp
cast monolithically with
the floor slab, ACI 11.5.1 defines the
overhanging flange width to be included in
the calculation of Acp and Pcp as shown
 The cracking torsion
Note: a lower limit (one fourth of derived) is placed for finding Tcr in the
previous equation for conservative purposes to deal with equilibrium
torsion. The derived value is permitted according to ACI code for
compatibility torsion.

Thin-walled hollow sections

The interaction diagram between shear and torsion approaches a straight
line as Ag /Acp decreases. Where Ag is the area of concrete only in a
cross section and Acp is the total area enclosed by the perimeter. As a
result, a value of T=0.25Tcr would reduce the inclined cracking shear to
0.75Vcr. So the ACI code replaces Acp in the previous equation of Tth
for solid section with Ag , so
 f c A 2g
T cr =
12 pcp
 Area of stirrups for torsion
Reinforced concrete beams subjected to torsion will have a decrease
in torsional strength after concrete cracks to about half of that of
the uncracked member, the remainder being now resisted by the
reinforcement. As the section approaches the ultimate load, the
concrete outside the stirrups cracks and begins to spall off. Thus,
the area enclosed by the dimensions xo and yo is now the one that
resists torsion. The xo and yo dimensions as shown in Figure a are
measured to the centerline of the outermost closed transverse
reinforcement and hence the gross area Aoh = xo yo and the shear
perimeter ph = 2(xo + yo).
Referring to Figure b, the
torsional resistance can be
represented as the sum of the
contributions of the shears in each
of the four walls of the equivalent hollow tube.
T n  T i  V y (x o /2)  V x ( y o/2)

Following a procedure similar to that used for shear, the quilibrium

of a section of the vertical wall with one edge parallel to a
torsional crack with angle θ can be evaluated
using the shown Figure as:

Atf yo
cot 
yt
V 2 = A t f yt n 
s
 Area of stirrups for torsion
Where: At = area of one stirrup leg of a closed stirrup.
f = yield strength of transverse reinforcement
yt
s = stirrup spacing.
Combining previous two
equations gives: 2Atf xo yo
T n=
yv
cot 
s
It has been found experimentally that, after cracking, the effective area
enclosed by the shear flow path is somewhat less than the value of xo yo
=Aoh , instead ACI recommends using 0.85 Aoh with Ao substituted for Aoh.
The value  may be taken between 30-60 degrees. ACI 11.5.3.6 suggests
that  may be taken as 45 degrees, because this corresponds to the angle
assumed in the derivation of the equation for designing stirrups for shear.
 Torsional longitudinal
reinforcement
The longitudinal reinforcement
Al must be proportioned to
resist the longitudinal tension forces that occur in the space truss.
A t f yt y o
N 2 V 2 cot   cot 2 
s
4
A t f yt A t f yt p h
N  N i  2(x o  y o ) cot  
2
cot 2   A l f y
i 1 s s
A t f yt p h
Al  cot 2

sf y
Tn A t f yt p h T p
or  A l  cot 2   n h cot 
2Ao At f yt cot  / s s f y 2Ao f y
Tu Tu /  p h
Ao  0.85Aoh ,T n   Al  cot 
 1.7 Aoh f y
 Combined shear and torsion

The previous derivations for stirrups and longitudinal

reinforcement in space truss analogy assumed all the torsion
was carried by the reinforcement without any torsion carried
by the concrete. When shear and torsion act together, the
1995 and subsequent ACI codes assume that:

Vn = Vc + Vs

Tn = Ts
 Maximum shear and torsion
A serviceability failure may occur if the inclined cracks are too wide at service loads.
The limit on combined shear and torsion in ACI 11.5.3.1 was derived to limit the
service-load crack widths.
For a hollow section as shown in Figure the two stresses add at A and yield:

V u experimental
T u p h results shows that above equation
For a member with a soildv section,
u =  2
is overconservative, a betterbrepresentation
w d 1.7 A oh is given by:

The combined shear stress due

to shear force and torsional
moment V 2 the same
T p
v uis=limited
( u to)  ( u 2h ) 2
bw dcapacity
criterion for shear 1.7as
A explained
oh in Chapter 6.
 Combined moment and torsion
• As shown in figure below: in the tension zone tensile forces
add together contrary to compression zone. ACI 11.5.3.9
allows the area in compression zone to be reduced by an
amount equal to M u /0.9dfy
 Design Procedure for Combined Shear,
Torsion and Moment
Step 1. Calculate the factored bending moment diagram or
envelope for the member.
Step 2. Select b, d, h and As based on Mu. Note: shallower wide-
beam sections are preferable to deep and narrow sections for
problems involving torsion.
Step 3. Given b and h, draw final Mu , Vu and Tu diagrams or
envelopes. Calculate the reinforcement required for flexure.
Step 4. for a solid section if  f c A 2cp
Tu 
12 pcp
Where φ=0.75, neglect torsion. Note: replaces Acp in the previous
equation with Ag for thin walled sections.
 Design Procedure for Combined Shear,
Torsion and Moment
 Step 5: If the torsion is compatibility torsion, the maximum factored
torque may be reduced to  f c A 2cp
At sections d from the faces of the supports (theT u moments and
3 pcp
shears in the other members must be adjusted accordingly).
Equilibrium torsion cannot be adjusted. 
Step 6: Check shear stresses in the section under combined torsion
and shear, for solid section if:
 
Vu 2 Tu p h 2 Vc 2
( ) ( 2
)  (  fc )
bw d 1.7A oh bw d 3
increase dimensions. The critical section for shear and torsion is
located a distance d from the face of the support. 
 Design Procedure for Combined Shear,
Torsion and Moment
For hollow sections
Vu Tu p h Vc 2
 2
 (  fc )
bw d 1.7A oh bw d 3
If the wall thickness varies, the equation is evaluated at the location where the
left-hand side is the greatest.
If a hollow section has a wall thickness, t, less than Aoh/ph , ACI 11.5.3.3 requires
that the actual wall thickness be used. Thus the second term in the above
equation becomes Tu /(1.7Aoh t)
Steps 7-9. Calculate the required transverse reinforcement for torsion and shear:
Compute Vs = Vu /Φ - Vc; then calc. Av/s=Vs/fyt d where fyt 4,20MPa.
If increase the size of the cross section.

2
V s> f c b wd
3
 Design Procedure for Combined Shear,
Torsion and Moment
Find: At = T n
s 2Ao f yt

Then combine shear and torsional transverse reinforcement for a

typical two-leg stirrup as:
Av t Av At
= 2
s s requirements:
Check minimum transverse reinforcement s

A v  2A t  0.062 f c b w s/ f yt , or (0.35b w s/ f yt )

Solve for the required spacing of closed stirrups s, and compare it

with ph/8 or 30cm maximum spacing for torsion (ACI 11.5.6.1)
and d/2 or d/4 maximum spacing for shear.
 … Design Procedure for Combined Shear
Step 10: Compute longitudinal area of steel using the larger of:
A t f yt p h
Al  cot 2 
sf y

5 f c Acp At f yt At 0.175bw
A l ,min   ph , 
12f y s f y s f yt
Satisfy the spacing (should not exceed 30cm), and bar size requirements (the
diameter of longitudinal bar may not be less than s/24 or 10mm). Torsion
reinforcement must be symmetrically distributed around all cross section
and that part which needs to be placed where As is needed must be
added to As found in step 1. Torsion reinforcement must be extended at
least a distance d+bt beyond the section where

f c A 2cp
Tu 
12 pcp
 Additional remarks
1. Fy 420MPa to limit crack widths ACI 11.5.3.4
2. The transverse stirrup used for torsional reinforcement
must be of a closed form. The concrete outside the
reinforcing cage is not well anchored, and the shaded region
will spall off if the compression in the outer shell is large as
shown in figure:
 Additional remarks
Thus ACI 11.5.4.2 (a) requires
that stirrups or ties must be
anchored with a 135o hooks
around longitudinal bars if
the corner can spall. ACI
11.5.4.2 (b) allows the use of
a 90 degrees standard hook
if the concrete surrounding
the anchorage is restrained
against spalling by a flange or
a slab.
 Additional remarks
3. If flanges are included in the computation of torsional
strength for T and L-shaped beams, closed torsional stirrups
must be provided in the flanges as shown in Figure.
 Example 7.2
A cantilever beam 1.35m long supports its own dead load plus a
concentrated load located 0.15m from the end of the beam and
0.15m away from the centroidal axis of the beam as shown. The
beam supports its own dead load plus an unfactored
concentrated load which is 90kN dead load and 90kN live load.
Design reinforcement for flexure,
shear, and torsion.
Use fy = 420MPa for all steel
and f’c = 21MPa.
 Example 7.3
Redesign the beam in the previous example using a hollow cross
section. Try the section shown in figure below.
 Example 7.4
The one-way joist system
shown in figure supports
A 16kN/m total factored load
applied directly to beam AB
including beam own weight.
The factored load on the slab
is 15kN/m2.
Design the end span AB of the exterior spandrel beam on grid line
1. Use fy = 420MPa for all steel and f’c = 28MPa.
 Homework 1
1. A cantilever beam 2.4m long and 450mm wide supports its own dead
load plus a concentrated load located 150mm from the end of the beam
and 115mm away from the vertical axis of the beam. The concentrated
load is 67kN dead load and 90kN live load. Design reinforcement for
flexure, shear, and torsion. Use fy = 420MPa for all steel and f’c = 26MPa.
2. Given the floor system shown in Fig. P7-3. f’c =315MPa, fy = 420MPa for
the longitudinal steel and fy = 280MPa for the transverse steel :
a) Design the spandrel beam between columns A1 and B1 for bending,
shear, and torsion. Check all of the appropriate ACI Code requirements for
strength, minimum reinforcement area, and reinforcement spacing are
satisfied.
•  
 
(b) Design the spandrel
beam between columns
A1 and A2 for bending,
shear, and torsion.
Check that all of the
appropriate ACI Code
requirements for
strength, minimum
reinforcement area, and
reinforcement spacing
are satisfied.
•