3.5 Analysis of Members under Flexure (Part IV)

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3.5 Analysis o Members under Flexure (Part IV) This section covers the ollowing topics. Analysis o a Flanged Section 3.5.1 Analysis o a Flanged Section Introduction A beam can have langes or lexural eiciency.

1 3.5 Analysis o Members under Flexure (Part IV) This section covers the ollowing topics. Analysis o a Flanged Section Analysis o a Flanged Section Introduction A beam can have langes or lexural eiciency. There can be several types o langed section. 1) A precast or cast-in-place langed section, with langes either at top or bottom or at both top and bottom. ) A composite langed section is made o precast web and cast-in-place slab. The ollowing igures show dierent types o langed sections. T-section Double T-section Single box section Double box section L-section Inverted T-section I-girder Figure Examples o precast langed sections

2 T T T T T T Box section Composite beam-slab T-section Figure 3-5. Examples o composite langed sections The analysis o a langed section or ultimate strength is dierent rom a rectangular section when the lange is in compression. I the depth o the neutral axis rom the edge under compression is greater than the depth o the lange, then the section is treated as a langed section. In the ollowing igure, the irst strain proile shows that the depth o the neutral axis (x u ) is greater than the depth o the lange (D ). The section is treated as a langed section. The second strain proile shows that x u is less than D. In this situation, the section can be treated as a rectangular section. b d D x u x u A p b w Cross-section Strain proile Strain proile (x u > D ) (x u < D ) Figure Two possibilities o strain proile in a langed section The eective width or breadth o the lange (b ) is determined rom the span o the beam, breadth o the web (b w ) and depth o the lange (D ) as per Clause 3.1., IS: Analysis o a Flanged Section The ollowing sketch shows the beam cross-section, strain proile, stress diagram and orce couples at the ultimate state. The ollowing conditions are considered. 1) x u > D : This requires an analysis or a langed section.

3 ) D (3/7) x u : This ensures that the compressive stress is constant at ck along the depth o the lange. b ck 0.5D D d x u 0.4x u C C u uw + A p ε p T uw T u b w Cross-section ε Strain Stress Force Figure Sketches or analysis o a langed section The variables in the above igure are explained. b b w D d A p ε p x u ε breadth o the lange breadth o the web depth o the lange depth o the centroid o prestressing steel (CGS) area o the prestressing steel strain dierence in the prestressing steel when strain in concrete is zero depth o the neutral axis at ultimate strain in prestressing steel at the level o CGS at ultimate stress in prestressing steel at ultimate The strain dierence ( ε p ) is urther explained in Section 3.4, Analysis o Member under Flexure (Part III). In the sketch, the tensile orce is decomposed into two components. The irst component (T uw ) balances the compressive orce carried by the web, including the portion o the lange above web (C uw ). Thus T uw C uw. The second component (T u ) balances the compressive orce carried by the outstanding portion o the lange (C u ). Thus T u C u. The stress block in concrete is derived rom the constitutive relationship or concrete. The relationship is explained in Section 1.6, Concrete (Part II). The compressive orce in concrete can be calculated by integrating the stress block along the depth. The

4 stress in the tendon is calculated rom the constitutive relationship or prestressing steel. The relationship is explained in Section 1.7, Prestressing Steel. The expressions o the orces are as ollows. C 0.36 x b uw ck u w u ck w uw u pw p (3-5.1) C (b - b )D (3-5.) T A T A (3-5.3) (3-5.4) The strengths o the materials are denoted by the ollowing symbols. A p A pw ck pk part o A p that balances compression in the outstanding langes part o A p that balances compression in the web characteristic compressive strength o concrete characteristic tensile strength o prestressing steel Based on the principles o mechanics (as explained under the Analysis o a Rectangular Section in Section 3.4, Analysis o Member Under Flexure (Part III)), the ollowing equations are derived. 1) Equations o equilibrium The irst equation states that the resultant axial orce is zero. This means that the compression and the tension in the orce couple balance each other. F 0 T C u u T +T C +C uw u uw u ( pw p ) ck u w ck ( w) A +A 0.36 x b b -b D (3-5.5) The second equation relates the ultimate moment capacity (M ur ) with the internal couple in the orce diagram. ( 0.4 ) ( 0.5 ) ( ) ( ) M ur Tuw d- x u +Tu d- D A d-0.4x +A d-0.5 D (3-5.6) pw u p From T u C u and Eqns. (3-5.) and (3-5.4), A p is given as ollows. The calculation o A pw rom A p and A p is also shown.

5 pw p p ( ) 0.447ck b -bw D A p A A -A (3-5.7) (3-5.8) ) Equation o compatibility The depth o the neutral axis is related to the depth o CGS by the similarity o the triangles in the strain diagram. x u d ε - ε p (3-5.9) 3) Constitutive relationships a) Concrete The constitutive relationship or concrete is considered in the expressions o C uw and C u. This is based on the area under the design stress-strain curve or concrete under compression. b) Prestressing steel ( ) F ε (3-5.10) The unction F(ε ) represents the design stress-strain curve or the type o prestressing steel used. The known variables in an analysis are: b, b w, D, d, A p, ε p, ck and pk. The unknown quantities are: A p, A pw, M ur, x u, ε and. The objective o the analysis is to ind out M ur, the ultimate moment capacity. The simultaneous equations to can be solved iteratively. The steps o the strain compatibility method are as ollows. 1) Assume x u D. ) The calculations are similar to a rectangular section, with b b. 3) I T u > C u, increase x u. Treat the section as a langed section. 4) Calculate ε rom Eqn. (3-5.9). 5) Calculate rom Eqn. (3-5.10). 6) Calculate A p and A pw rom Eqn. (3-5.7) and Eqn. (3-5.8), respectively.

6 7) Calculate C uw, C u, T uw and T u rom Eqns. (3-5.1) to (3-5.4). I Eqn. (3-5.5) (T u C u ) is not satisied, iterate with a new value o x u, till convergence. 8) Calculate M ur rom Eqn. (3-5.6). The capacity M ur can be compared with the demand under ultimate loads. In the strain compatibility method, the diicult step is to calculate x u and. Similar to the rectangular section, an approximate analysis can be done based on Table 11 and Table 1, Appendix B, IS: The tables are reproduced in Table and Table 3-4., respectively, in Section 3.4, Analysis o Member under Flexure (Part III). The values o x u and are available in terms o a reinorcement index ω pw. Apwpk ω (3-5.11) bd pw w ck Note that the index is calculated based on A pw instead o A p. The calculation o A pw is rom Eqn. (3-5.8). But A p depends on, which is unknown. Hence, an iterative procedure is required. The steps are as ollows. 1) Assume 0.87 pk. ) Calculate A p and A pw rom Eqn. (3-5.7) and Eqn. (3-5.8), respectively. 3) Calculate ω pw. 4) Calculate rom Table 11 or Table 1. Compare the calculated value o with the assumed value. Repeat steps 1 to 4 till convergence. 5) Calculate M ur. I D > (3/7) x u, the lange depth is larger than the depth o constant compressive stress. An equivalent depth o the lange is deined as ollows. y 0.15x u D (3-5.1) The equivalent depth y is substituted or D in the expression o M ur.

7 Example A bonded post-tensioned concrete beam has a langed cross-section as shown. It is prestressed with tendons o area 1750 mm and eective prestress o 1100 N/mm. The tensile strength o the tendon is 1860 N/mm. The grade o concrete is M60. Estimate the ultimate lexural strength o the member by the approximate method o IS: Values are in mm. Cross-section at mid-span Solution Eective depth d mm Assume x u D 175 mm. Treat as a rectangular section, with b b 460 mm. Reinorcement index ω P A bd P Pk ck

8 T u A p kn C u 0.36 ck x u b kn T u > C u. Hence x u > D Treat as a langed section Assume 0.87 pk 1618 N/mm Calculate A p and A pw A p ck ( b - bw ) D ( ) mm A pw mm Reinorcement index ω pw A bd pw pk w ck From Table 11, 0.87 pk N/mm

9 nd iteration 1489 N/mm Calculate A p and A pw A p ( ) mm A pw mm Reinorcement index ω pw From Table 11, 3 rd iteration 0.87 pk N/mm 151N/mm Calculate A p and A pw A p A pw ( ) mm mm Reinorcement index The value o w pw is same as ater nd iteration. Hence, the values o, A p and A pw have converged. ω pw

10 Ultimate lexural strength ( ) ( ) M T d-0.4 x + T d-0.5d ur uw u u T ( d- 0.4 x ) A ( d- 0.4 x ) uw u pw u ( ) knm Tu ( d-0.5 D ) Ap ( d-0.5 D ) ( ) knm The ultimate lexural strength is given as ollows. M ur knm

9.5 Compression Members

9.5 Compression Members This section covers the following topics. Introduction Analysis Development of Interaction Diagram Effect of Prestressing Force 9.5.1 Introduction Prestressing is meaningful when