1
Analysis and Design Analysis and Design for Torsionfor Torsion
Dr. Husam Al Qablan
for Torsionfor 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
lt f
Dr. Husam Al Qablan
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. the stressing of adjacent members as the beam twists permit a
Dr. Husam Al Qablan
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).
Dr. Husam Al Qablan
2. Floor system: equilibrium torque (circular beams).
3. Circular tie beams in mosques.
Torsional effects in reinforced concrete
Dr. Husam Al Qablan Dr. Husam Al QablanCompatibility Torsion
2
Dr. Husam Al Qablan
Shearing Stresses Due to Torsion in Un-cracked Members
Dr. Husam Al Qablan
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
Dr. Husam Al Qablan
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,
Dr. Husam Al Qablan
J πr /2 its polar moment of inertia, and τmax is the maximum elastic shearing stress due to elastic twisting moment T, then from basic strength (mechanics) of material courses
maxrT=
Jτ
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 (τcorners=0) and the maximum shear stresses occur at the midpoints of the long sides as
Dr. Husam Al Qablan
shown in Figure. These complications plus the fact that reinforcedconcrete sections are neither homogeneous nor isotropic make it difficult to develop exact mathematical formulations based on the physical models.
Dr. Husam Al Qablan
3
Dr. Husam Al Qablan
Hollow membersConsider a thin-wall tube subjected to a torsion T as shown in the Fig. below. 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:
2AB 1 1 2 2 1 1 2CD= V dx = dx = qV t t t tττ τ τ→ → =
Dr. Husam Al Qablan
Where q is referred to as the shear flow and is constant.
Hollow membersIn 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
PT = rqdx∫
Dr. Husam Al Qablan
but rds = twice the area of the shadedtriangle, then
where Ao is the area enclosed by the middle of the wall of the tube. From the above equation τmax occurs where t is the least.
P q∫
2 oo
q TT = q =A t 2A tτ→ =
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 I i f d h k ld
Dr. Husam Al Qablan
In reinforced concrete such a crack would Cause failure unless it was crossed by reinforcement. This generally takes the formof longitudinal bars in the corners and closedstirrups.
Principal stresses due totorsion and shear
The two shear stresses components add on oneside face (front side) and counteract each otheron the other. As the result inclined cracking starts on AB and extends across the flexural
Dr. Husam Al Qablan
tensile face. If bending moments are large, the cracks will extend almost vertically across theback face. The flexural compression zone near the bottom prevents the cracks from extending full height.
Behavior of RC members subjected to torsionWhen a concrete member is loaded inpure torsion, shear stresses develop. One or more cracks (inclined) developwhen the maximum principal tensile stress reaches the tensile strength of conc ete The onset of c acking
Dr. Husam Al Qablan
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.
4
Behavior of RC members subjected to torsionAfter 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
Dr. Husam Al Qablan
y p p p p yof 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.
Space Truss AnalogyTheory
Assumptions:1. Both solid and hollow members
are considered as tubes.2. After cracking the tube is
idealized as a hollow truss
Dr. Husam Al Qablan
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 θ, generally taken as 45 degrees for RC.
The cracking pure torsionKnowing 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
ti A /P h P i th i t f th ti Th i
o
q T=t 2A t
τ =
Dr. Husam Al Qablan
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:
2
3cpc
crcp
AfT =
p
According to ACI-R11.6.1
)(2;:
43;
32
o
cpcp
cp
cpcpo
tandAofvaluesngsubstituti
yxPxyAwhere
PA
tAA
+==
==
Dr. Husam Al Qablan
4
31 2
'
cru
cp
cpccr
TT
whenTorsionNeglect
PA
fT
φ≤
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛=⇒
Vertical Stirrups and Longitudinal steel Reinforcement
Steel area in one leg stirrup leg is
oyv
ut Af
sTA
φ2=
Dr. Husam Al Qablan
oyvfφ
Longitudinal steel reinforcement
yo
hul fA
PTAφ2
=
Combined Shear and Torsion
dbV
w=τ
AT
2=τ
Shear Stress :
Torsional Stress :
Dr. Husam Al Qablan
tAo2
hohoho pAtAA /;85.0 ==For cracked section :
5
Dr. Husam Al Qablan
27.1 oh
h
wtv
ATp
dbV
+=+= τττ2
2
2
7.1 ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
oh
h
w ATp
dbVτ
ACI Requirement for Torsional Design
Equilibrium Torsion: Design for full uT Compatibility Torsion: reduce uT to the following
Nonprestressed member without a ial force
Dr. Husam Al Qablan
Nonprestressed member without axial force
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
cp
cpc P
Af
2'
31φ
Nonprestressed member with axial force
'
2'
33.01
31
cg
u
cp
cpc
fA
NPA
f +⎟⎟
⎠
⎞
⎜⎜
⎝
⎛φ
Neglect Torsion effect if the factored torsional moment is less than Nonprestressed member without axial force
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ cpc P
Af
2'083.0 φ
Dr. Husam Al Qablan
⎠⎝ cpP
Nonprestressed member with axial force
'
2'
33.01083.0
cg
u
cp
cpc
fA
NPA
f +⎟⎟
⎠
⎞
⎜⎜
⎝
⎛φ
The cross sectional dimensions shall be such that For solid section
⎟⎟⎠
⎞⎜⎜⎝
⎛+≤⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= '
2
2
2
max 66.07.1 c
w
c
oh
hu
w
u fdb
VApT
dbV
φτ
Dr. Husam Al Qablan
For hollow section
⎟⎟⎠
⎞⎜⎜⎝
⎛+≤+= '
2max 66.07.1 c
w
c
oh
hu
w
u fdb
VApT
dbV
φτ
If NOT increase section dimensions
Reinforcement for torsion Recall ACI Eq (11-21)
oot TA
Dr. Husam Al Qablan
oo
oyv
utAfT
sA 6030;
cot2≤≤= θ
θφ
Combined shear and torsion reinforcement
sA
sA
sA tvtv 2
+=+
Maximum spacing of torsion reinforcement
⎪⎨⎧ p
ofsmallersh
8
Dr. Husam Al Qablan
⎪⎩⎨=
mmofsmallers
3008max
Spacing is limited to ensure the development of the ultimate torsional strength of the beam, to prevent excessive loss of torsional stiffness after cracking, and to control crack widths.
6
Minimum area of closed stirrups
( )⎪⎪
⎪⎪
⎨
⎧
=+wc
yv
w
tv sbf
fsb
oferlAA '062.0
35.0
arg2
Dr. Husam Al Qablan
⎪⎩ yvf
Minimum area of longitudinal torsional reinforcement
yv
wt
y
vth
t
y
cpcl
fb
sAwhere
ffp
sA
fAf
A
175.0
42.0 '
min,
≥
⎟⎠⎞
⎜⎝⎛−=
The longitudinal reinforcement shall be distributed around the perimeter of the closed stirrups with a maximum spacing of 300 mm
Dr. Husam Al Qablan
mm The longitudinal bars shall be inside the stirrups The longitudinal bars Shall have a diameter at least 0.042 times the stirrups spacing but not less than Ø10
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: for problem involving torsion, square cross-sections are
Dr. Husam Al Qablan
p ob e o g to s o , squa e c oss sect o s a epreferable.
Step 3. Given b and h, draw final Mu , Vu and Tu diagrams or envelopes. Calculate the reinforcement required for flexure.
Step 4. Determine whether torsion must be considered. Torsion must be considered if Tu exceeds the torque given by
⎟⎟⎞
⎜⎜⎛ cpA
f2
'0830 φ
Design Procedure for Combined Shear, Torsion and Moment
Dr. Husam Al Qablan
Otherwise, it can be neglected.
Where φ=0.75
⎟⎟
⎠⎜⎜
⎝ cpc P
f083.0 φ
Design Procedure for Combined Shear, Torsion and Moment
Step 5: If the torsion is compatibility torsion, the maximum factored torque may be reduced to
2
3cpc
ucp
AfT
pφ
≤
Dr. Husam Al Qablan
At sections d from the faces of the supports (the moments and shears in the other members must be adjusted accordingly). Equilibrium torsion cannot be adjusted.
cp
Step 6: Check shear stresses in the section under combined torsion and shear, for solid section
⎟⎟⎞
⎜⎜⎛
+≤⎟⎟⎞
⎜⎜⎛
+⎟⎟⎞
⎜⎜⎛
= '2
2
2
max 66.0 cchuu f
VpTVφτ
Dr. Husam Al Qablan
The critical section for shear and torsion is located a distance d from the face of the support.
⎟⎠
⎜⎝⎟
⎠⎜⎝
⎟⎠
⎜⎝
2max 7.1 cwohw
fdbAdb
φ
7
Design Procedure for Combined Shear, Torsion and Moment
Steps 7-9. Calculate the required transverse reinforcement for torsion and shear:
Compute Vs = Vu /Φ - Vc; then calc. Av/s=Vs/fyt dwhere fyt ≤4,20MPa.
Dr. Husam Al Qablan
If increase the size of the
cross section.
23 cs w> dV bf
Design Procedure for Combined Shear, Torsion and Moment
Find:
Then combine shear and torsional transverse reinforcement for a typical two-leg stirrup as:
t n
yto
A T=s 2 fA
Dr. Husam Al Qablan
reinforcement for a typical two leg stirrup as:
Check minimum transverse reinforcement requirements:
2v t v tA A A =s s s
+ +
2 0.062 , (0.35 )v t yt ytw wcf s/ or s/b bA A f f+ ≥
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.
Dr. Husam Al Qablan
Design Procedure for Combined Shear …Step 10: Compute longitudinal area of steel using the larger
of:
S ti f th i ( h ld t d 30 ) d b i
2cott yt hl
y
pA fA s f
θ=
,min5 0.175,
12c cp ytt t w
l hy y y
f A A A bfpA s s tf f f= − ⋅⋅⋅ ≥
Dr. Husam Al Qablan
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
2
12cpc
ucp
AfT
p≤
Additional remarks
1. Fy ≤420MPa to limit crack widths ACI 11.5.3.42. The transverse stirrup used for torsional
reinforcement must be of a closed form. The concrete outside the reinforcing cage is not well
Dr. Husam Al Qablan
co c ete outs de t e e o c g cage s ot eanchored, and the shaded region will spall off if the compression in the outer shell is large as shown in figure:
Additional remarksThus ACI 11.5.4.2 (a) requires that stirrups or ties must be anchored with a 135o hooks around longitudinal bars
Dr. Husam Al Qablan
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.
8
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.
Dr. Husam Al Qablan
Example: Torsion
wtownunfactoredkNPkNP
MpaffMpaf
LL
DL
yvy
c
+⎭⎬⎫
=
=
==
=′
8585
42020
Dr. Husam Al Qablan
Assume
Dr. Husam Al Qablan
mkNP
mkNwt
mkNwt
ult
factored
/238856.1852.1
/91.676.52.1
/76.5246.04.0
=×+×=
=×=
=××=
Dr. Husam Al Qablan
Moments at the support
mkNT
mkNM
/8.352/15.00365.115.0238
/5.3302/5.15.191.635.10365.135.1238
=×+×=
=××+×+×=
Dr. Husam Al Qablan
Design for Flexure
Assume mmda 75.1334/ ==
( ) ( )
,
26
3.1154002085.0
4207.186785.0
7.1867102/75.1335354209.0
5.3302/
mmbf
fAa
mmadf
MA
c
ys
y
us
=××
×==
=×−×
=−Φ
=
Dr. Husam Al Qablan
( )
2
26
1831
min
max
1831102/3.1155354209.0
5.330
mmA
Aforcheck
Aforcheck
forcheck
mmA
s
s
s
t
s
=
=×−×
=
ε
9
Should Torsion be Considered?
( )
Af
mmP
mmA
cpc
cp
cp
⎟⎞
⎜⎛⎟
⎞⎜⎛
=+=
=×=
24000020
20006004002
240000600400
622,
2
Dr. Husam Al Qablan
consideredbemustTorsionT
mkNP
fT
u
cp
cpc
⇒>=
=×⎟⎟⎠
⎜⎜⎝
×=⎟⎟
⎠⎜⎜
⎝Φ= −
05.88.35
.05.8102000
240000122075.0
126
θ
The torsion is needed for equilibrium
⇒ Design for 8.35=uT
Is the section large enough to resist torsion?
⎟⎟⎠
⎞⎜⎜⎝
⎛+≤⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= '
2
2
2
max 66.07.1 c
w
c
oh
hu
w
u fdb
VApT
dbV
φτ
1
Dr. Husam Al Qablan
dbfV wcc'
61
=
ohA = area within centerline of closed stirrups
Assume 40mm cover mm10Φ
Dr. Husam Al Qablan
( ) mmyxPmmymmx
mmxyA
h
oh
1640)25101040260031010402400
158100310510 2
=+==−×−==−×−=
=×==
MPadb
MPa
MPa
w795220660
2061
75081
8.11581007.1
1640108.35535400107.245
2
2
623
=⎟⎟⎞
⎜⎜⎛
+≤
=⎟⎟⎠
⎞⎜⎜⎝
⎛
×
××+⎟
⎟⎠
⎞⎜⎜⎝
⎛
××
⇒
Dr. Husam Al Qablan
MPadb
MPaw
795.22066.075.08.1 =⎟⎟⎟
⎠⎜⎜⎜
⎝
+≤
The cross section is large enough
Compute the stirrups area required for shear
( )scu VVV +Φ≤
kNdbfV wcc 5.159105354002011 3' =×××== −
Dr. Husam Al Qablan
mmmms
A
dfV
sA
ors
dfAV
kNV
f
v
y
svyvs
s
wcc
/748.0535420101.168
1.1685.15975.0
7.24566
23
=××
=⇒
==
=−=
For shear ⇒ require stirrups with mmmms
Av /748.0 2=
Compute the stirrup area required for torsion
TA
mmNT
oont
n
6030
.107.4775.0
108.35 66
≤≤
×=×
=
θ
Dr. Husam Al Qablan
mmmms
A
mmAA
Afs
t
ooho
oo
oyv
nt
/422.04201343852
107.47
45
13438515810085.085.0
6030;cot2
26
2
=××
×=
=
=×=×=
≤≤=
θ
θθ
10
Add the stirrup area and select stirrups
mmmms
AsA
sA
sA
tv
tvtv
/59.1422.02748.0
2
2=×+=
+=
+
+
Ch k i i ti
Dr. Husam Al Qablan
Check minimum stirrups
( )
⎪⎪⎩
⎪⎪⎨
⎧
=+
yv
w'c
yv
w
tv
fbf
161
fb
31
oferargls
A2A
OK
mm/mm266.0420
40020161
mm/mm317.0420400
31
sA
min
2
2
tv
⇒
⎪⎪
⎩
⎪⎪
⎨
⎧
=×
=
⇒ +
Dr. Husam Al Qablan
Use ⇒10Φ Area for two legs = 2mm157
mm7.98s59.1s
15759.1s
A tv =⇒=⇒=+
Use ⇒12Φ Area for two legs = 2mm226
mm2.142s =
Check maxs
OKmm300
mm2058
16408p
ofsmallersh
max ⇒⎪⎩
⎪⎨⎧ ===
⇒ Use 12Φ closed stirrups at 125 mm on center
Design the longitudinal reinforced for torsion
2yvfA ⎟⎞
⎜⎛
⎟⎞
⎜⎛
Dr. Husam Al Qablan
22l
t
2
yl
yvh
tl
mm69245cot4204201640422.0A
422.0s
A
cotff
ps
AA
=⎟⎠⎞
⎜⎝⎛××=
=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛= θ
yl
yvh
t
y
cp'c
min,l ff
psA
fAf
125A ⎟
⎠⎞
⎜⎝⎛−=
OKmmmmAl ⇒<=××−×
= 22min, 692372
4204201640422.0
42024000020
125 we need 6
bars < 30 mm spacing
Each area 2mm33.1156/692A =≥
Dr. Husam Al Qablan
The min bar dim {
⎪⎩
⎪⎨⎧
≥
=×=
10
25.5125042.0
φ
mmspacingstirrup
Provide 144φ in the bottom half of the beam and add
22
mm76414*4692 =⎟⎟
⎠
⎞⎜⎜⎝
⎛−
π to the flexural steel
2mm1907761831A +
Dr. Husam Al Qablan
s mm1907761831A =+=
Example: Compatibility Torsion
The one-way joist system shown in the figure below supports a
total factored dead load of 2/5.7 mkN and factored live load of 2/8 mkN , totaling 2/5.15 mkN . Design the end span AB, of the
exterior spandrel beam on the grid line 1 the factored dead load
Dr. Husam Al Qablan
exterior spandrel beam on the grid line 1. the factored dead load of the beam and the factored loads applied directly to it total
mkN /16 . The spans and loading are such that the moments and the shears can be calculated by using the moment coefficients from ACI section 8.3.3. Use Mpafc 30=′ ,. Mpaff yvy 420==
11
Dr. Husam Al Qablan
Compute the bending moments for the beam.
In laying out the floor, it was found that joist with an overall depth of 470 mm would be required. The slab thickness is 110 mm. the spandrel beam was made the same depth, to save forming costs. The columns
Dr. Husam Al Qablan
supporting the beams are 600 mm square. For simplicity in forming the joist, the beam overhangs the inside face of the columns by 50 mm. thus, the initial choice of the beam size is h = 470 mm, b = 650 mm, and d = 405 mm.
The joist reaction per meter of length of beam is
mkNmmkNwln /1.722
30.9/5.152
2=
×=
Th t t l l d th b i
Dr. Husam Al Qablan
The total load on the beam is
mkNw /1.88161.72 =+=
Using mmln 6600= . The moments in the edge beam are as follows
Exterior end negative mkNwlM nu /9.239
16
2−==
Midspan positive mkNwlM nu /1.274
14
2==
First interior negative mkNwlM nu /8.383
10
2−==
The areas of steel required for flexure are as follows:
Dr. Husam Al Qablan
Exterior end negative: 21791mmAs =
Midspan positive: 22046mmAs =
First interior negative: 22865mmAs =
The actual steel will be chosen when the longitudinal
torsion reinforcement has been calculated.
Compute the final moment shear and
Dr. Husam Al Qablan
moment, shear, and torsion diagrams
The exterior negative moment in the joist
mkNwlM nu /9.55
243.95.15
24
22−=
×==
Although this is a bending moment in the joist,
it acts as a twisting moment on the edge beam.
Dr. Husam Al Qablan
g g
As shown in the figure below, this moment and
the shear of 72.1 kN/m act at the face of the edge beam.
The torque at the center of the beam = t = 79.3 kN.m/m
81.5kN.m/m the torque transferred to the col
12
If the two ends of the beam A-B are fixed
against rotation by the column,
the total torque at each end will be
mkNtlT n .7.2612
6.63.792
=×
==
The total shear at end A of the beam will be
66188 ×
Dr. Husam Al Qablan
kNdV
kNV
u
u
2552
6.61.88@
7.2902
6.61.88
=×
=
=×
=
The total shear at end B of the beam will be
kNdV
kNV
u
u
7.2982
6.61.88@
3.3342
6.61.8815.1
=×
=
=×
×=
Dr. Husam Al Qablan
Dr. Husam Al Qablan Dr. Husam Al Qablan
Should torsion be considered?
consideredbemustTorsionT
mkNPAf
T
mmP
mmA
u
cp
cpc
cp
cp
⇒>=
=×⎟⎟⎠
⎞⎜⎜⎝
⎛×=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛Φ=
=+++++=
=×+×=
−
8.137.261
.8.13102960
345100123075.0
12
29601010110360360650470
345100360110650470
622,
2
θ
Dr. Husam Al Qablan
Equilibrium or compatibility torsion?
The torque resulting from the 25-mm offset of the axes
of the beam and column is necessary for equilibrium torque.
The torque at the ends of the beam due to this is
mkN .3.726.6025.01.88 =××⇒
On the other hand, the torque resulting from
the moments at the ends of the joists exist only
because the joint is monolithic and the edge beam
has a torsional stiffness. If the torsional stiffness
were to decease to zero, this torque would disappear.
Dr. Husam Al Qablan
This part of the torque is therefore compatibility torsion.
mkN
PAf
dTcp
cpcu
.1.55102960
34510033075.0
3@
62
2,
=×⎟⎟⎠
⎞⎜⎜⎝
⎛×=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛Φ=⇒
−
But not less than the equilibrium torque
13
Assuming the remaining torque after redistribution
is evenly distributed along the length of the spandrel
beam. The distribution reduced torque t, due to moments
at the ends of the joists has decreased to
mkNt .192/)405.026.6(
1.55=
×−=⇒
Dr. Husam Al Qablan
Adjust the moments in the joists
Because the analysis procedure (ACI moment
coefficients) assumed an exterior support (spandrel
beam) with zero torsional stiffness,
No load or moment redistribution would be required.
Is the section large enough to resist torsion?
⎟⎟⎠
⎞⎜⎜⎝
⎛+≤⎟
⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= '
2
2
2
max 66.07.1 c
w
c
oh
hu
w
u fdb
VApT
dbV
φτ
dbfV wcc'
61
=
mmp
mmAoh
1868)577377(2
209989577377 2
=+=
=×=
Dr. Husam Al Qablan
mmph 1868)577377(2 =+=
MPadb
dbMPa
MPa
w
w423.33066.0
3061
75.08.1
781.12099897.1
1868101.55405650107.298
2
2
623
=⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
+≤
=⎟⎟⎠
⎞⎜⎜⎝
⎛
×
××+⎟
⎟⎠
⎞⎜⎜⎝
⎛
××
⇒
The cross section is large enough
Compute the stirrups area required for shear in the edge beam
kNdbfV wcc 313.240104056503061
61 3' =×××== −
At the left of the beam (End B)
2076.1405420
24031375.0
103.334/
3
=×
−×
=−
==df
VVdf
Vs
Ay
cu
y
sv φ
At d f d B
Dr. Husam Al Qablan
At d from end B
9286.0405420
24031375.0
107.298/
3
=×
−×
=−
==df
VVdf
Vs
Ay
cu
y
sv φ
Compute the stirrups required for torsion
42020998985.0210/;
cot2/ 6
××××
==φ
θφ u
oyv
ut TAf
Ts
A
At end B, 5583.0.8.62 =⇒=s
AmkNT tu
At d from end B, 4902.0.1.55 =⇒=s
AmkNT tu
At d from end A, 4902.0.1.55 =⇒=s
AmkNT tu
Dr. Husam Al Qablan
Add the stirrup area and select stirrups
909.14902.029286.0
2
=×+=
+=
+
+
sA
sA
sA
sA
tv
tvtv
Provide No. 13M closed stirrups
End A: one @75 mm, seven @150 mm
End B: one @75 mm, 12@125 mm, then @200 mm
on centers through the rest of the span
Design the longitudinal reinforcement for torsion
2cotff
ps
AA yvh
tl ⎟
⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛= θ
Dr. Husam Al Qablan
29161118684902.0 mmA
fs
l
yl
=×××=
⎟⎠
⎜⎝⎠⎝
2min,
,
min,
960
118684902.042012345100305
125
mmA
ff
ps
AfAf
A
l
yl
yvh
t
yl
cpcl
=
××−××
=⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛−=
Use 2960mmAl =
To satisfy ACI requirements, we need 3 bars
at the top and bottom and one halfway up each side
22 1208/960/ mmmmbarAs ==
Exterior end negative moment
2215112031791 mmA =×+=
Dr. Husam Al Qablan
215112031791 mmAs =×+=
Use 8 No. 19M
First interior negative moment
2322512032865 mmAs =×+=
Use 7 No. 25M
14
Midspan positive moment
2240612032046 mmAs =×+=
Use 5 No. 25M
Dr. Husam Al Qablan Dr. Husam Al Qablan
Dr. Husam Al Qablan Dr. Husam Al Qablan