Q1: An R.C.C beam of 25 cm width has a clear span of 5 metres and carries a U.D.L. of 2000 kg/m inclusive of its self weight. If the lever arm of the section is 45 cm., the beam is

ANS:A - safe in shear

To determine whether the RCC beam is safe under the given conditions, we need to check its shear capacity. The maximum shear force (Vmax​) that occurs in the beam can be calculated using the factored load (w) and the clear span (L): Vmax​=wL​/2 Given:

• Factored load (w) = 2000 kg/m
• Clear span (L) = 5 meters

Vmax​=2000×5​ / 2 =5000kg Now, let's compare this maximum shear force (Vmax​) with the shear capacity of the beam. The shear capacity (Vc​) of an RCC beam without stirrups can be calculated using the formula: ​=0.87×fc​​×b×d Where:

• fc​ is the characteristic compressive strength of concrete
• b is the width of the beam
• d is the effective depth of the beam

Given:

• Width of the beam (b) = 25 cm = 0.25 m
• Effective depth of the beam (d) = Lever arm = 45 cm = 0.45 m (as provided)

Now, let's assume the characteristic compressive strength of concrete (fc​) to be 20 N/mm² (approximately 200 kg/cm²). Vc​=0.87×200​×0.25×0.45 Vc​≈0.87×14.14×0.25×0.45 Vc​≈0.87×14.14×0.1125 Vc​≈1.1135kg/cm2 So, the shear capacity of the beam (Vc​) is approximately 1.1135 kg/cm². Comparing the maximum shear force (Vmax​) with the shear capacity of the beam (Vc​): =5000 kgVmax​=5000kg =1.1135 kg/cm2Vc​=1.1135kg/cm2 Since Vmax​ exceeds Vc​, the beam is unsafe in shear. Therefore, the correct conclusion is: The beam needs revision of the section.