- 2079 Bhadra · 6 marks
Derive moment capacity of doubly reinforced WSM
Short answer
Using 1.5m for compression steel, neutral axis is bx^2/2+(1.5m-1)Asc(x-d')=mAst(d-x). Moment of resistance is Mr=0.5σcbc bx(d-x/3)+(1.5m-1)Asc σcbc[(x-d')/x](d-d'). Assumptions include plane sections, no tension in concrete, and perfect bond.
Derive equations for bending moment carrying capacity of a double reinforced rectangular section using assumptions for working stress method given by IS 456:2000.
Assumptions (IS 456:2000, Annex B-1.2)
- Plane sections remain plane after bending.
- Concrete is elastic in compression and carries no tension.
- Perfect bond between steel and concrete; stress in steel modular ratio stress in surrounding concrete.
- For compression steel, effective modular ratio is (to allow for creep).
Notation: = width, = effective depth, = effective cover to compression steel, = depth of neutral axis, = tension steel, = compression steel, = modular ratio.
Section Strain Stresses/forces
+------+ |\ Cc (concrete)
d'| Asc | | \ x Cs (comp. steel)
| | -+--- NA
| | | /
| Ast | |/ T (tension steel)
+------+
Stress in concrete at the level of compression steel
By similar triangles of the strain diagram:
Stress in compression steel . The concrete displaced by the steel is allowed for by using as the equivalent concrete area.
Neutral axis (first moment of the equivalent area about the neutral axis is zero)
Moment of resistance
Taking moments about the tension steel, with concrete compression acting at from the top:
Substituting :
The first term is the moment of the singly reinforced (balanced) part when , and the second is the additional moment from the compression steel. The tension steel then follows from .
Checks: , and (190 N/mm for Fe415).
✓ Answer checked by Claude Opus 5.5 (high effort)
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