Competitive Exam practice-1

Two rectangular under-reinforced concrete beam sections X and Y are similar in all aspects except that the longitudinal compression reinforcement in section Y is 10% more.  Which one of the following is the correct statement?

• Section X has less flexural strength and is less ductile than section Y

• Section X has less flexural strength but is more ductile than section Y

•   Section X and Y have equal flexural strength but different ductility

• Section X and Y have equal flexural strength and ductility

Answer explanation

Both the sections i.e., X and Y are under reinforced. So, in both the cases the flexural strength (MOR) can be calculated from tension side.

When additional compression reinforcement is placed i.e., in case of cross-section Y, neutral axis will shift upward.

 

MOR = Tension X Lever arm

In case of cross section Y, lever arm is more as depth of neutral axis is less. However, tensile force is constant in both cases as the steel would have yielded on both cases. Hence flexural capacity of ‘Y’ is more.

Also, as section Y is more strong in compression, yielding in steel will be more at limit state of collapse

As per IS456:2000, the minimum percentage of tension reinforcement (up to two – decimal places) required in reinforced-concrete beams of rectangular cross-section (considering effective depth in the calculation of area) using Fe500 grade steel is 

• 0.06

• 0.17

• 0.25

• 0.8

Answer explanation

A_st/bd = 0.85/f_y

A_st/bd = 0.85/500

Percentage = A_st/bd X 100

                  = 0.85/500 X 100 = 0.17

The singly reinforced concrete beam section shown in the figure (not drawn to the scale) is made of M25 grade concrete and Fe500 grade reinforcing steel. The total cross-sectional area of the tension is 942mm². As per limit state design of IS456:2000, the design moment capacity (in kNm round off to two decimal places) of the beam section is       

• 162.57 kN-m

• 158.27 kN-m

• 168.27 kN-m

• 152.57 kN-m

Answer explanation

M25; Fe500

A_st = 942mm²

X_ulimit = 0.46d = 0.46 x 450 = 207mm

For calculating of depth of NA

0.36 f_ck b X_u = 0.87 f_y A_st

Ø 0.36 x 25 x 300 x X_u = 0.87 x 500 x 942

Ø X_u = 151.76mm < X_ulimit

So, under reinforcement section.

Moment of resisting for under reinforced section.

MR = 0.36 f_ck b X_u (d-0.42X_u)

       = 0.36 x 25 x 300 x 151.76(450-0.42x151.76)

       = 158271135.3 N-mm

       = 158.27kNm

The cross-section of the reinforced concrete beam having an effective depth of 500mm is shown in the figure (not drawn to the scale). The grades of concrete and steel used are M35 and Fe550, respectively.  The area of tension reinforcement is 400mm².  It is given that the corresponding to 0.2% proof stress, the material safety factor is 1.15 and the yield strain of Fe550 steel is 0.0044. As per IS 456:2000, the limiting depth (in mm, round off to nearest integer) of the neutral axis measured from the extreme compression fiber, is

Answer explanation

From similar triangulation

X_ulimit/0.0035 = d- X_ulimit/0.0044

X_ulimit / (d- X_ulimit) = 0.79545

X_ulimit = 0.44304d

X_ulimit = 0.44304 x 500

              = 221.52

              = 222mm

An RCC beam of rectangular cross-section has factored shear of 200kN at its critical section. Its width b is 250mm and effective depth is 350mm. Assume design shear strength τc of concrete as 0.62 N/mm2 and maximum allowable shear stress τcmax in concrete as 2.8N/mm2. If two legged 10mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per limit state method will be

• 5.5

• 8.2

• 15.6

• 19.4

Answer explanation

Given, V_u = 200kN, τ_c = 0.62MPa

τ_cmax = 2.8MPa

τ_v = V_u/bd = 200 x 103/250 x 350 = 2.286MPa

As τ_v < τ_cmax and design shear force = (τ_v – τ_c)bd

= (2.286 – 0.62) x 250 x 350

=145.775kN

Spacing of shear reinforcement calculation:

V_us = 145775kN

=0.87 x 250 x 2 x π/4 x 102 x 350/S_v

S_v = 82.03mm

Spacing for minimum shear reinforcement

A_sv/bS_v ≥ 0.4/0.87f_y

S_v ≤ 0.87f_yA_sv/0.4b

S_v≤341.65mm

Spacing should be minimum of

i)  0.75d = 26.25cm

ii) 8.2cm

iii) 34.16cm

iv) 30cm

So, spacing will be 8.2cm