Unformatted text preview: 3.4.3 Torsional shear stress – A plastic method of analysis
In BS 8110: Part 2, clause 2.4.4, a plastic analysis is recommended for
calculating the torsional shear stress vt and ultimate torsional moment T,
where the expressions are derived from the sandheap analogy.
• The sand heap for a rectangular section is shown in Fig. 3.46, where hmin and hmax are the smaller and larger dimensions of the
section, respectively. a Fig. 3.46 Sand heap for rectangle section
• In the sandheap model shown in Fig. 3.46 from theory of plasticity,
o Torsional shear stress vt = tan ψ ; thus vt = 2a / hmin (3.34) o Ultimate torsional moment T = twice the volume of the sand heap; thus 1
⎡1 2
⎤
T = 2 ⎢ hmin a + hmin a (hmax − hmin )⎥
2
⎣3
⎦ (3.35) • From Eq. (3.34), a = vt hmin / 2, and Eq. (3.35) then becomes
2
T = (vt hmin / 2) ⋅ ( hmax − hmin / 3) (3.36) 146 ο From Eq. (3.6), the torsional shear stress of a rectangular section is given by vt = 2
min h 2T
( hmax − hmin / 3) (3.37) ο For a thin hollow section, vt = T /(2 Aht )
where A is the enclosed by centreline of the wall, and ht is the
wall thickness.
• Sandheaps of flanged sections Fig. 3.47 Approximate sand heaps for flanged sections
o For a flanged section, it should be divided into component rectangles. The torque resisted by each rectangle
3
(hmin hmax )i
Ti = T
3
∑ ( hmin hmax ) (i = 1,2,…n) (3.38) o It is seen from Eq. (3.38) that the division of a flanged section is to be such as to maximise the function of ∑ (h 3
min ) hmax . This can be achieved if the widest rectangle is made as long as
possible. 147 Example; Division of a flanged section
1600
150
500
300 Arrangement 1 Arrangement 2 (1) For Arrangement 1: Rib = 500 × 300, Flange = 650 × 150 ∑ (h 3
min )( )( ) hmax = 300 3 × 500 + 2 × 150 3 × 650 = 17.89 × 10 9 (2) For Arrangement 2: Rib = 350 × 300, Flange = 1600 × 150 ∑ (h 3
min max h ) = (300 3 )( ) × 350 + 1503 ×1600 = 14.85 × 10 9 ∴ The Tsection should be divided into the component rectangles
shown in the Arrangement 1.
• Torsional reinforcement must be provided if the torsional shear stress vt exceeds the torsional capacity of the concrete section.
o The limiting value of the torsional capacity of concrete sections recommended by BS 8110 is expressed as vt,min, given by
vt,min = 0.067√fcu , but not more than 0.4 N/mm2.
o Values of vt,min are also given in Table 3.41.
Table 3.41 Torsional shear stresses (N/mm2)
(BS 8110: Part 2, clause 2.4.5)
Concrete grade
25 30 40 or higher vt,min 0.33 0.37 0.40 vtu 4.00 4.38 5.00 (Allowance is made for γm) 148 3.4.4 Torsion combined with direct shear and bending
Torsion is seldom present alone, and will normally be combined with
direct shear and bending stresses in most practical situations.
• Combined with direct shear ⎛ o The sum of the shear stress from direct shear ⎜ v = ⎝ V⎞
⎟ and
bv d ⎠ torsional shear stress vt (Eq. 3.37) must not exceed the ultimate
torsional shear stress vtu , v + vt ≤ v tu (3.39) where vtu ≤ 0.8 f cu or 5 N/mm2, whichever is the lesser. Values
of vtu for different grades of concrete are given in Table 3.41.
o Additionally in the case of small sections where y1 < 550 mm, vt v tu y1
550 (3.40) This condition is to prevent spalling of the corners of sections.
• Combined with bending Fig. 3.48 Interaction curve for combined bending and torsion 149 o As flexural cracks only slightly reduce the torsional stiffness, for the moment up to approximately 0.8Mu , the section can still
resist the full ultimate torsion Tu , as shown in Fig. 3.48.
o Hence, no calculation for torsion is generally necessary for the ultimate state of reinforced concrete unless torsion has been
included in the original analysis or is required for equilibrium.
3.4.5 Design for torsion
The procedure of design for torsion of a rectangular section is
summarised as follows.
1. Calculate the torsional shear stress using Eq. (3.37), vt =
where T = 2
min h 2T
(hmax − hmin / 3) torsional moment due to ultimate load hmax = larger dimension of the section
hmin = smaller dimension of the section
2. If the torsional shear stress vt > vt,min = 0.067√fcu with an upper limit of 0.4 N/mm2 (see page 148 Table 3.41), torsional reinforcement
must be provided.
3. Reinforcement for direct shear and torsion Links may be required for direct shear and/or torsion, and may
be provided according to the following guidelines: 150 Table 3.42 Reinforcement for direct shear (v) and torsion (vt )
v t > v t,min v t ≤ v t,min 4. If (v + vt ) > vtu = 0.8√fcu with an upper limit of 5 N/mm2 (see page 148 Table 3.41), design is inadmissible. The crosssection dimensions must then be made larger.
For the case of small sections (y1 < 550 mm), vt should not exceed vtu y1
y
, i.e. vt ≤ vtu 1 .
550
550 5. Calculate the required torsional reinforcement in the form of closed links using Eq. (3.32), Asv
T
≥
sv 0.8 ⋅ x1 y1 (0.95 f yv )
The requirements of reinforcement detailing are given in the
Section 3.4.3 of the Notes page 145, particularly
sv ≤ min {200 mm, x1, y1
}, whichever is the least.
2 6. Calculate the torsional longitudinal reinforcement required using Eq. (3.33),
As ≥ Asv f yv
(x1 + y1 )
sv f y 151 o This additional As should be distributed evenly around the inside perimeter of the links.
o At least four bars should be used and the clear distance between bars 300 mm 6. A flanged section should be divided into component rectangles and
each component designed to carry a torsional moment given by
3
( hmin hmax )i
Ti = T
3
∑ (hmin hmax ) Example: Design for torsional reinforcement
The rectangular section shown in the following figure resists a bending
moment of 170 kNm, a shear force of 160 kN and a torsional moment of
10 kNm. The characteristic material strengths are fcu = 30 N/mm2, fy =
460 N/mm2 and fyv = 250 N/mm2. Calculations for bending and direct
shear would give As = 1000 mm2 and Asv
= 0.73 . Design for torsional
sv reinforcement. 152 1. Torsional shear stress vt =
= 2
min h 2T
(hmax − hmin / 3) 2 × 10 × 106
= 0.56 N/mm2
2
300 × (500 − 300 / 3) 2. From Table 3.41, vt,min = 0.37.
vt = 0.56 > vt,min = 0.37. Thus torsional reinforcement is required.
3. The ultimate torsional shear stress
vtu = 0.8√fcu = 0.8√30 =4.38 N/mm2 (or directly from Table 3.41).
The value y1 = 440 mm < 550 mm; then the section is a small section.
Check:
vt = 0.56 N/mm2 < vtu y1
440
= 4.38 ×
= 3.5 N/mm2 as required.
550
550 The direct shear stress is
v= V
160 × 103
=
= 1.19 N/mm 2 ;
bv d 300 × 450 v + vt = 1.19 + 0.56 = 1.75 N/mm2 < vtu = 4.38 N/mm2 OK 4. Links for torsional shear stress Asv
T
=
sv 0.8 ⋅ x1 y1 (0.95 f yv )
10.0 ×106
=
0.8 × 240 × 440 × 0.95 × 250
= 0.50
Therefore,
Total stirrup ratio for direct shear
for torsional shear Asv
= 0.73 + 0.50 = 1.23
sv
153 Provide R10 links at 100 mm centres ([email protected]), Asv
= 1.57
sv
The links are of the closed type with their ends fully anchored.
5. Torsional longitudinal steel As = Asv f yv
⋅
( x1 + y1 )
sv f y 250
× (240 + 440)
460
= 185 mm 2
= 0.50 × 6. Reinforcement detailing 2T12 [email protected] 2T12 185/3 mm2
185/3 mm2 1000 + 185/3 mm2 ο The torsional longitudinal bars should extend at least hmax beyond where it is required to resist the torsion. 154 ...
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 Spring '06
 JSKUANG
 Force, Shear Stress, Polar moment of inertia, torsional shear

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