General Engineering (Civil and Structural): design theory and IS code values
This domain tests the theory behind the calculations: structural analysis, mechanics of materials, reinforced concrete and steel design, and concrete technology. SSC JE expects IS-code-aware answers, so know which method you are in (limit state or working stress) before touching a formula, and quote partial safety factors correctly.
Core rules
- Limit State Method: design stress in steel is 0.87 fy; the concrete stress block is 0.36 fck b xu, with xu/d limited to 0.48 for Fe415 and 0.46 for Fe500.
- Bending: limiting moment Mu,lim = 0.138 fck b d squared for Fe415, and 0.148 fck b d squared for Fe500.
- Shear: nominal shear stress is Vu / (b d); design shear strength depends on the percentage of tension steel (IS 456 Table 19); the maximum shear stress is about 0.62 times the square root of fck for M20 to M40.
- Deflection control by span/depth ratio: 20 for simply supported, 26 for continuous, 7 for cantilever beams.
- Steel: design shear strength of a bolt derives from fub divided by root 3 and by the partial safety factor 1.25 for bolts.
Worked example
A singly reinforced beam has b = 230 mm, d = 450 mm, M20 concrete and Fe415 steel. Find the limiting moment. Mu,lim = 0.138 x 20 x 230 x 450 squared = 0.138 x 20 x 230 x 202,500 = 128.6 kN-m. If the applied factored moment is 100 kN-m the section is under-reinforced and needs only tension steel. If it is 150 kN-m the section is over-reinforced and requires compression steel or a larger depth.
Traps
Using fy directly without the 0.87 factor; applying the Fe415 xu/d limit to Fe500 sections; mixing working stress and limit state partial safety factors in one calculation; assigning the cantilever span/depth ratio of 7 to a continuous slab; and assuming design shear strength depends only on concrete grade when it also depends on steel percentage.