Essay: Concrete calculations

Equation is valid provided the neutral axis is in the flange, dn < t.
The force C then acts at a depth
If the neutral axis is in the web of a T-section, the force C has to be reduced by an amount Cn given by which can be regarded as a negative force acting at a depth
Above equations allow the internal forces and hence M, to be calculated from the strain distribution in the section i.e., from εo and dn.
Usually in a cracked section analysis, M is known and the strain and stresses have to be determined by trial and error.
It is convenient to start with a trial value of dn and find the corresponding value of εo.
At all stages of loading, the longitudinal force equilibrium must be satisfied.
C – Cn=Tp+Ts
By substituting expressions already derived for the various forces, an equation relating εo to dn derived by Warner is useful for computations and is expressed as follows;
This equation allows εo to be determined for any chosen value of dn in the range D>dn>t.
For rectangular sections, bw must be replaced by b.
If the neutral axis is in the flange (dn < 1),
Above Eq. can be used provided the term in the denominator containing (b – bw) is set to zero and bw is set to b.
It is important to note that the present analysis assumes a linear elastic behaviour and that inelastic behaviour in either concrete or steel may develop as the neutral axis rises into the flange.
After evaluating the trial value of dn, the internal forces can be evaluated and the moment corresponding to dn can be obtained.
M = Tpdp + Tsds +Cndzn – Cd2
Usually, the computations are made for two appropriate dn values of 0.5D and 0.35D.
Resulting values of M will usually bracket the working moment, which can then be found by interpolation. Entire sequence of calculations is sufficiently small to be programmed on a modem calculator.
According to the European concrete committee, the width of a crack is related to the stress in the reinforcement.
In the case of partially prestressed beams, the F.I.P, BS: 8110 and IS: 1343 codes limit the width of cracks under service loads to 0.1 mm in an aggressive environment and to 0.2 mm in normal situations.
British and Indian codes provide for the limitation of crack width by computation of hypothetical flexural tensile stresses as detailed in Table.
Investigations by Parameswaran have indicated that, the hypothetical tensile stress method is suitable for predicting with reasonable accuracy, the maximum width of cracks in class-3 type members.
Use of the approximate method involving the stress and strain compatibility of the section for computing the width of cracks are illustrated by the following examples.
Let us discuss about the design of a partially prestressed post tensioned beam (class 3 type) to suit the following data:
Effective span = 30 m.
Live load = 9 kN/m.
Dead load (excluding self weight) = 2 kN/m.
Load factors = 1.4 for dead load and 1.6 for live load
28-day cube compressive strength, fcu = 50 N/mm2.
Strength of concrete at transfer,.fci = 35 N/mm2.
Loss ratio, η = 0.85.
Tensile strength of concrete = 1.7 N/mm2.
Permissible tensile stress under service loads = 6 N/mm2.
Maximum width of crack under service loads not to exceed 0.1 mm. 8 mm diameter high tensile wires having an ultimate tensile strength of 1500 N/mm2 are available for use.
Design calculations for a partially prestressed beam are similar with regard to:
1.Ultimate moments and shear forces.
2.Cross-sectional dimensions.
3.Properties of section.
4.Design moments and shear forces.
Permissible stresses in concrete at the stage of transfer and service loads are as follows:
Maximum permissible eccentricity, e = 580 mm.
Magnitude of the prestressing force is 75 per cent of that required for the fully prestressed beam.
Four Freyssinct cables containing 12 wires of 8 mm diameter stressed to 1000 N/mm2, each providing a force of 600 kN, are provided.
Force developed in tendons at the limit state of collapse
Assuming the concrete compression to be located at the center of flange thickness.
But the ultimate moment required, Mud= 3400 kN m
Hence, balance moment = (3400 – 3050) = 350 kN m
If A – area of untensioned reinforcement made up of deformed bars (fy = 420 N/mm2), which arc provided at the soffit at a cover of 30 mm, then
Aus(0.87fy) (1300 – 125 – 30) = 350 × 106
∴Aus = 830 N/mm2
Five, 20 mm diameter high-yield bars are sufficient for providing the required area as well as for limiting the width of cracks.
Asu- 1570 mm2
Check for stresses
Stage
Prestress
Stresses due to loads
Resultant stresses (N/mm2)
Mmin
Md
At transfer
Top (6.6 – 11) = -4.4
7.8
–
+3.4
Bottom (6.6 + 14.1) = 20.7
-10.0
–
+10.7
At service loads
Top (0.85) × (-4.4) = -3.8
–
17.8
+140.0
Bottom (0.5 × 20.7) = 17.5
–
-22.5
-5.0
The stresses are within the permisssible limits.
Check for width of cracks
According to the provisions of the British code BS: 8110 – 1985 and IS: 1343 – 1980, the maximum permissible tensile stress in post-tensioned beams for a limiting crack-width of 0.1 mm using M-50 grade concrete and with one per cent untensioned reinforcement expressed as a percentage of the cross-sectional area of the concrete in the tension zone is given as,
0.7(4.8 + 4.0) = 6.16 N/mm2
Hypothetical flexural tensile stress at the soffit under service loads is only 5 N/mm2.
Hence, the beam is safe against excessive crack-width under service loads.
Let us consider the cross-section of a class-3 type post tensioned T-girder, designed to resist a service load moment of 1560 kN m. The beam is prestressed by a cable containing 19 strands of 12.7 mm diameter stressed to 1133 N/mm2. The supplementary reinforcements comprise six bars of 24 mm diameter. Using the rigorous method of cracked section analysis, estimate the width of cracks developed in the beam under the working moment.
Cross Section of Partially Prestressed T-Girder
(a) Properties of Section:
Ac = (160 × 1200) + (250 × 640) = 352000 mm2.
Ap = (19 × 98.7) = 1875 mm2.
Ic = 1983 × 107 mm4.
Es = Ep = 200 kN/mm2 Ec = 31.6 kN/mm2.
Effective stress in tendons = 1133 N/mm2.
Prestressing force, P = (1875 × 1133) = 2124 × 103 N.
Eccentricity, e = 238 mm.
(b) Strains in Tendons and Concrete:
(c) Cracking Moment:
Compressive prestress at soffit,
Modulus of rupture of concrete,
= 902 × 106 N mm.
= 902 kNm.
(d)Computation of εo for Trial dn:
Assuming a trial dn = 250 mm and εce = 0.00032.
εpe = 0.0057
bw = 250 mm
Ap = 1875 mm2
Ep = 200 kN/mm2
b = 1200 mm
dp = 500 mm
As = 2713 mm2
t = 160 mm
ds = 700 mm
Ec = 31.6 kN/mm2
(e)Forces and Moment:
dn = 250 mm
εo = 0.00076
εpe=0.0057
εce=0.00032
Since the computed value of the moment M is nearly equal to the service load moment of 1560 kN m, the assumed value of dn = 250 is correct.
If not, a new trial value of dn is assumed and the computation procedure is repeated until the moment values are very nearly equal.
(f) Computation of Crack Widths:
(i) According to the British code BS: 8110-1985, the design surface crack-width at the soffit of the girder directly under the bar is
wmax = 3 Cmin εm
Where ,Cmin = Clear cover.
εm = Average strain at the level where cracking, is considered.
Neglecting the stiffening effect of concrete in the tension zone, the strain at soffit is
∴Cmin = 38 mm
wmax = (3 × 38 × 0.00167) = 0.182 mm
(ii)According to Nawy’s method outlined in Section 11.7, the maximum crack-width at the level of the centroid of tension steel is
whereAt = (250 × 362) = 90500 mm2.
Σ0 = (19 × 39.25) + (6 × 75.36) = 1198 mm.
∆fs = increase of stress in steel from the decompression stage
Maximum crack-width at the tensile face of the beam
(g)Hypothetical Flexural Tensile Stresses and Permissible Crack Width:
According to the British and Indian standard codes, the hypothetical flexural tensile stresses, corresponding to a maximum permissible crack-width of 0.1 to 0.2 mm, depends upon the supplementary steel in the tension zone.
Compressive prestress at soffit = 19.7 N/mm2.
Tensile stress at soffit, corresponding to the service load moment of M = 1560 kN m is
Tensile stress at soffit = (42.39 – 19.7) = 22.69 N/mm2.
For fck = 50 N/mm2, the maximum permissible tensile stress for post-tensioned grouted beams for a crack width of 0.2 mm is = (5.8) + (4 × 1.35)= 11.2 N/mm2.
Since the actual tensile stress exceeds the permissible value, the width of cracks will exceed 0.2 mm.
Limitations of Prestressing:
Although prestressing has advantages, some aspects need to be carefully addressed. They are:
• Prestressing needs skilled technology. Hence, it is not as common as reinforced concrete.
• Use of high strength materials is costly.
• Additional cost in auxiliary equipments.
• Need for quality control and inspection.