Question 1 |

A reinforced concrete beam with rectangular cross section (width = 300 mm,
effective depth = 580 mm) is made of M30 grade concrete. It has 1%
longitudinal tension reinforcement of Fe 415 grade steel. The design shear
strength for this beam is 0.66 N/mm^2. The beam has to resist a factored shear
force of 440 kN. The spacing of two-legged, 10 mm diameter vertical stirrups of
Fe 415 grade steel is ______mm. (round off to the nearest integer)

127 | |

101 | |

254 | |

331 |

Question 1 Explanation:

b=300mm

d=580 mm

V_u=440 kN

Concrete used is M30

Raft steel is Fe415

V_{cu}=\tau _c Bd=0.66 \times 300 \times \frac{580}{1000}=114.84 kN

V_{su}=V_u-V_{cu}=440-114.84=325.16 kN

Spacing of 2-legged shear reinforcement

\begin{aligned} s_V&=\frac{A_{SV} \times 0.87 f_y \times d}{V_{su}}\\ &=\frac{2 \times \frac{\pi}{2} \times (10)^2 \times 0.87 \times 415 \times 580}{325.16 \times 1000}\\ &=101.16 mm \end{aligned}

d=580 mm

V_u=440 kN

Concrete used is M30

Raft steel is Fe415

V_{cu}=\tau _c Bd=0.66 \times 300 \times \frac{580}{1000}=114.84 kN

V_{su}=V_u-V_{cu}=440-114.84=325.16 kN

Spacing of 2-legged shear reinforcement

\begin{aligned} s_V&=\frac{A_{SV} \times 0.87 f_y \times d}{V_{su}}\\ &=\frac{2 \times \frac{\pi}{2} \times (10)^2 \times 0.87 \times 415 \times 580}{325.16 \times 1000}\\ &=101.16 mm \end{aligned}

Question 2 |

A reinforcing steel bar, partially embedded in concrete, is subjected to a tensile force P. The figure that appropriately represents the distribution of the magnitude of bond stress
(represented as hatched region), along the embedded length of the bar, is

A | |

B | |

C | |

D |

Question 3 |

Tie bars of 12 mm diameter are to be provided in a concrete pavement slab. The working tensile stress of the tie bars is 230 MPa, the average bond strength between a tie bar and concrete is 2 MPa, and the joint gap between the slabs is 10 mm. Ignoring the loss of bond and the tolerance factor, the design length of the tie bars (in mm, round off to the nearest integer) is _____

350 | |

700 | |

1050 | |

850 |

Question 3 Explanation:

L_d=\frac{\phi \sigma _{st}}{4\tau _{bd}}=\frac{12 \times 230}{4 \times 2}=345mm

Length of tie bar =L_d+10+L_d=345+10+345=700mm

Question 4 |

For a given loading on a rectangular plain concrete beam with an overall depth of 500 mm, the compressive strain and tensile strain developed at the extreme fibers are of the same magnitude of 2.5 \times 10^{-4}. The curvature in the beam cross-section (in m^{-1}, round off to 3 decimal places), is ____

0.008 | |

0.006 | |

0.004 | |

0.001 |

Question 4 Explanation:

Simple bending equation

\begin{aligned} \frac{m}{I}&=\frac{f}{y}=\frac{E}{R} \\ \frac{1}{R}&=\frac{f}{E}\cdot \frac{1}{y} \\ &=\frac{\varepsilon }{y}=\frac{2.5 \times 10^{-4}}{(0.5/2)}=0.001m^{-1} \end{aligned}

\begin{aligned} \frac{m}{I}&=\frac{f}{y}=\frac{E}{R} \\ \frac{1}{R}&=\frac{f}{E}\cdot \frac{1}{y} \\ &=\frac{\varepsilon }{y}=\frac{2.5 \times 10^{-4}}{(0.5/2)}=0.001m^{-1} \end{aligned}

Question 5 |

An RCC beam of rectangular cross section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strength \tau _{c} of concrete as 0.62 N/mm^{2} and maximum allowable shear stress \tau _{c,max} in concrete as 2.8 N/mm^{2} . If two legged 10 mm 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 |

Question 5 Explanation:

Nominal shear stress,

\begin{aligned} \tau_{v}&=\frac{V_{u}}{b d}=\frac{200 \times 10^{3}}{250 \times 350}=2.286 \mathrm{N} / \mathrm{mm}^{2} \lt \tau_{c m a x} \\ &\text{(OK)}\\ SF&\text{ taken by stirrups}\\ &=\left(\tau_{v}-\tau_{c}\right) b d \\ &=(2.286-0.62) \times 250 \times 350 \\ &=145.75 \mathrm{kN} \\ \text{Now, }V_{u s}&=\frac{0.87 f_{y} A_{s v} d}{S_{v}} \\ \Rightarrow \quad S_{v}&=\frac{0.87 \times 250 \times 350 \times 2 \times \frac{\pi}{4} \times 10^{2}}{145.75 \times 10^{3}} \\ &=82 \mathrm{mm}=8.2 \mathrm{cm} \end{aligned}

\begin{aligned} \tau_{v}&=\frac{V_{u}}{b d}=\frac{200 \times 10^{3}}{250 \times 350}=2.286 \mathrm{N} / \mathrm{mm}^{2} \lt \tau_{c m a x} \\ &\text{(OK)}\\ SF&\text{ taken by stirrups}\\ &=\left(\tau_{v}-\tau_{c}\right) b d \\ &=(2.286-0.62) \times 250 \times 350 \\ &=145.75 \mathrm{kN} \\ \text{Now, }V_{u s}&=\frac{0.87 f_{y} A_{s v} d}{S_{v}} \\ \Rightarrow \quad S_{v}&=\frac{0.87 \times 250 \times 350 \times 2 \times \frac{\pi}{4} \times 10^{2}}{145.75 \times 10^{3}} \\ &=82 \mathrm{mm}=8.2 \mathrm{cm} \end{aligned}

Question 6 |

As per IS 456-2000 for the design of reinforced concrete beam, the maximum allowable shear stress \tau _{cmax} depends on the

grade of concrete and grade of steel | |

grade of concrete only | |

grade of steel only | |

grade of concrete and percentage of reinforcement |

Question 7 |

In shear design of an RC beam, other than the allowable shear strength of concrete (\tau _{c}) there is also an additional check suggested in IS 456-2000 with respect to the maximum permissible shear stress (\tau _{cmax}). The check for (\tau _{cmax}) is required to take care of

additional shear resistance from reinforcing steel | |

additional shear stress that comes from accidental loading | |

possibility of failure of concrete by diagonal tension | |

possibility of crushing of concrete by diagonal compression |

Question 8 |

The development length of a deformed reinforncement bar can be expressed as (1/k) (\phi \sigma _{s}/\tau _{bd}). From the IS: 456-2000, the value of k can be calculated as________.

6.4 | |

9.6 | |

3.4 | |

8.3 |

Question 8 Explanation:

Bond strength of concrete

\begin{aligned} &=\text{ Tensile force in steel}\\ \tau_{b d} \times\left(L_{d} \pi \phi\right)&=0.87 f_{y} \times\left(\frac{\pi}{4} \phi^{2}\right)\\ \Rightarrow \quad L_{d}&=\frac{0.87 f_{y} \phi}{4 \tau_{b d}} \\ \end{aligned}

For deformed bars \tau_{\text {bd }} value is increased by 60 %

\begin{aligned} \therefore \quad L_{d}&=\frac{\phi \sigma_{s}}{4 \times\left(1.6 \tau_{b d}\right)}=\frac{\phi \sigma_{s}}{6.4 \tau_{b d}}\\ \therefore \quad k&=6.4 \end{aligned}

\begin{aligned} &=\text{ Tensile force in steel}\\ \tau_{b d} \times\left(L_{d} \pi \phi\right)&=0.87 f_{y} \times\left(\frac{\pi}{4} \phi^{2}\right)\\ \Rightarrow \quad L_{d}&=\frac{0.87 f_{y} \phi}{4 \tau_{b d}} \\ \end{aligned}

For deformed bars \tau_{\text {bd }} value is increased by 60 %

\begin{aligned} \therefore \quad L_{d}&=\frac{\phi \sigma_{s}}{4 \times\left(1.6 \tau_{b d}\right)}=\frac{\phi \sigma_{s}}{6.4 \tau_{b d}}\\ \therefore \quad k&=6.4 \end{aligned}

Question 9 |

A rectangular beam of width (b) 230 mm and effective depth (d) 450 mm is reinforced with four bars of 12 mm diameter. The grade of concrete is M20 and grade of steel is Fe500. Given that for M20 grade of concrete the ultimate shear strength, \tau _{uc} = 0.36 N/mm^{2} for steel percentage, p = 0.25, and \tau _{uc}= 0.48 N/mm^{2} for p = 0.50. For a factored shear force of 45 kN, the diameter (in mm) of Fe500 steel two legged stirrups to be used at spacing of 375 mm, should be

8 | |

10 | |

12 | |

16 |

Question 9 Explanation:

\begin{aligned} \tau_{u c}=& 0.36 \mathrm{N} / \mathrm{mm}^{2} &[\text{For M20}]\\ & \text { for } \frac{A_{s t}}{b d} \times 100=0.25\\ \tau_{u c}=& 0.48 \mathrm{N} / \mathrm{mm}^{2} &[\text{For M20}]\\ & \text { for } \frac{A_{s t}}{b d} \times 100=0.5 \end{aligned}

Factored S F=45 \mathrm{kN}=V_{u}

We have to calculate the dia of Fe 500 2-Legged stirrup to be used at a spacing of 325 mm c/c

\begin{aligned} \tau_{v}&=\frac{V_{u}}{b d}=\frac{45 \times 1000}{230 \times 450}=0.4348 \mathrm{N} / \mathrm{mm}^{2}\\ \therefore \quad \%&\text{tensile steel}\\ &=\frac{4 \times \frac{\pi}{4}(12)^{2}}{230 \times 450} \times 100=0.437 \% \\ \tau_{c} &=0.36+\frac{0.12}{0.25} \times(0.437-0.25) \\ &=0.45 \mathrm{N} / \mathrm{mm}^{2} \end{aligned}

since \tau_{v}-\tau_{c} \lt 0

\Rightarrow Min shear reinforcement is required

\Rightarrow Min shear reinforcement is given by

\begin{aligned} \frac{A_{s v}}{b S_{v}} &=\frac{0.4}{0.87 f_{y}} \\ A_{s v} &=\frac{0.4 \times\left(S_{v}\right)(b)}{0.87 f_{y}} \end{aligned}

since we limit f_{y} to 415 \mathrm{N} / \mathrm{mm}^{2} hence,

\begin{aligned} A_{s v}=2 \times \frac{\pi}{4}(\phi)^{2} &=\frac{0.4 \times 325 \times 230}{0.87 \times 415} \\ &= 82.814 \mathrm{mm}^{2} \\ \phi &=7.26 \mathrm{mm} \\ \text { adopt } \phi&=8 \mathrm{mm} \end{aligned}

Question 10 |

As per IS 456:2000 for M20 grade concrete and plain barsin tension the design bond stress \tau _{bd}=1.2 MPa.Further, IS 456:2000 permits this design bond stress value to be increased by 60% for HSD bars. The stress in the HSD reinforcing steel barsin tension, \sigma _{s}=360MPa. Find the required development length, L_{d}, for HSD barsin terms of the bar diameter, \phi. __________

34.54 | |

46.87 | |

89.25 | |

93.47 |

Question 10 Explanation:

\begin{aligned} L_{d} &=\frac{\phi \sigma_{s t}}{4 \tau_{b d}}=\frac{\phi \times 360}{4 \times 1.2 \times 1.60} \\ &=46.875 \phi \end{aligned}

There are 10 questions to complete.