RCC Design Test 2

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RCC Design Test 2

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SHARING IS CARING

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Weekly Quiz Competition

Question 1
2 / -0.33

For the working stress method, limiting depth of Neutral axis depends on ____

A
Grade of concrete

B
Grade of steel

C
Percentage of steel

D
All of the above

Solution

Limiting depth of the neutral axis depends only on the grade of steel for Working stress method:
\(\rm{\begin{array}{l}\rm{ Limiting\;depth, }\;{x_u} = \frac{{m\;{\sigma _{cbc}}}}{{m\;{\sigma _{cbc}} + {\sigma _{st}}}}\\ x_u = \frac{1}{{1 + \left( {\frac{{{\sigma _{st}}}}{{m\;{\sigma _{cbc}}}}} \right)}}\\ x_u= \frac{1}{{1 + \left( {\frac{{{\sigma _{st}}}}{{\left( {\frac{{280}}{{3\;{\sigma _{cbc}}}}} \right) \times {\sigma _{cbc}}}}} \right)}}\\ x_u= \frac{1}{{1 + \frac{{3{\sigma _{st}}}}{{280}}}} \end{array}}\)
Where σ_cbc = permissible stress in concrete
σ _st = Permissible stress in steel
So the depth of the neutral axis only depends on the grade of steel.
Question 2
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While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/mm²) in limit state method of design as per IS : 456 - 2000 is __________

A
9-9

B
94-91

C
97-96

D
99-92

Solution

Concept
As per IS 456: 2000, the bearing strength of concrete (f_br) is given by:
f_br = 0.45 f_ck (For limit state)
Calculation
Given,
Grade of concrete M20 i.e. f_ck = 20 MPa
∴ f_br = 0.45 × 20 = 9 MPa
Question 3
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An axially loaded column is of 300 mm × 300 mm size. The effective length of the column is 3 m. What is the minimum eccentricity of the axial load for the column?

A
25 mm

B
10 mm

C
16 mm

D
20 mm

Solution

As per IS 456: 2000, clause 25.4,
Minimum Eccentricity
All columns shall be designed for minimum eccentricity, equal to the addition of the unsupported length of column divided by 500 and lateral dimensions divided by 30, subject to a minimum of 20 mm.
Where biaxial bending is considered, it is sufficient to ensure that eccentricity exceeds the minimum about one axis at a time.
Calculation:
Unsupported length = 3000 mm
Size of the column = 300 mm
Minimum eccentricity = \(\frac{L}{{500}} + \frac{B}{{30}} \)
\(e_{min}= \;\frac{{3000}}{{500}} + \frac{{300}}{{30}} = 16\;mm\;\)
But, in no case, the minimum eccentricity should be less than 20 mm.
Hence, the minimum eccentricity of the axial load for the column is 20 mm.
Question 4
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A reinforced cement concrete beam section has a size of 300 mm width and 400 mm effective depth, the grade of concrete used is M20 and grade of steel is Fe 415. If the ultimate bending moment of 150 kN-m acts at the section, the beam has to be designed as

A
Singly reinforced beam

B
Doubly reinforced beam

C
Beam cannot be designed

D
Balanced beam

Solution

Approach: Check the ultimate and limiting bending moment.
Concept:
To know which type of beam need to be used is determined by checking Ultimate bending moment and Limiting Bending moment.
Need for Doubly reinforced reinforcement
i) When depth is restricted
ii) Reversal of stresses.
iii) Subjected to Impact loads
iv) M_U> M_Ulim
\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = {\rm{compressive\;force}} \times {\rm{lever\;arm\;\;}}\)
\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = {\rm{C}} \times \left( {{\rm{d}} - 0.42{{\rm{x}}_{{\rm{umax}}}}} \right){\rm{\;\;}}\)
\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = 0.36{\rm{\;}}{{\rm{f}}_{{\rm{ck}}}} \times {\rm{b}} \times {{\rm{x}}_{{\rm{umax}}}} \times \left( {{\rm{d}} - {{\rm{x}}_{{\rm{umax}}}}} \right){\rm{\;}}\)
Where,
Limiting bending moment (M_Ulim)
Ultimate bending moment (M_U)
d is the effective distance
Steel
X_Umax
Fe250
0.53×d
Fe415
0.48×d
Fe500
0.46×d

Calculation:
Given Grade of concrete M20
Width of beam = 300mm
Depth of the beam = 400mm
Grade of steel is Fe415
Ultimate bending moment (M_U) = 150kN-m
Limiting bending moment (M_Ulim)
\({{\rm{M}}_{{\rm{Ulim}}}}{\rm{\;}} = 0.36{\rm{\;}}{{\rm{f}}_{{\rm{ck}}}} \times 0.48 \times {\rm{d}} \times \left( {{\rm{d}} - 0.48 \times {\rm{d}}} \right){\rm{\;}}\)
\({M_{Ulim}} = 0.138{f_{ck}} \times b \times {d^2}\) kN-m
\({M_{Ulim}} = 0.138 \times 20 \times 300 \times {400^2}\)
\(= 132.48\) kN-m
∴ M_U > M_Ulim
∴ Doubly reinforced beam is used.
Question 5
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In a reinforced concrete section, the stress at the extreme fibre in compression is 12 MPa. The depth of neutral axis in the section is 60 mm and grade of concrete is M20.

Assume linear elastic behavior of the concrete.

A
Radius of curvature is 11.16 cm

B
Effective curvature is 8.96 × 10⁵mm^-1

C
Effective curvature is 8.96 × 10⁵mm

D
Radius of curvature is 11.16 m

Solution

Concept

Bending Equation is given by

\(\frac{M}{I} = \frac{E}{R} = \frac{\sigma }{y}\)

Where, M is moment, σ is flexural stress

I is moment of inertia, y is distance from neutral axis to extreme fibre.

1/R is curvature.

R is radius of curvature

Modulus of Rigidity is given by E = 5000 × √f_ck

Where f_ck is characteristic strength of concrete.

Calculation

Given,

y = 60 mm, σ = 120 MPa,

Grade of concrete = M20 i.e. fck = 20 N/mm²

Modulus of Elasticity of concrete E = 5000× √f_ck

E = 5000 × √20

= 22,360.68 N/mm²

From the bending equation

\(\frac{M}{I} = \frac{E}{R} = \frac{\sigma }{y}\)

Where, 1/R is curvature.

R is radius of curvature

σ is flexural stress

\(\frac{1}{R} = \frac{\sigma }{{E \times y}} = \frac{{12}}{{22,360.68 \times 60}} = 8.94 \times {10^{ - 5}}\;m{m^{ - 1}}\)

∴ Effective Curvature or Curvature = 8.94 × 10^-5per mm

Radius of Curvature = \(R = \frac{{E \times y}}{\sigma }\)

= 10⁵/8.94 = 11,160.71 mm

= 11.16 m
Question 6
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If ϕ = nominal diameter of reinforcing bar, f_s= compressive stress in the bar and f_bd = design bond stress of concrete,
Which of the following statements are true for development length (L_d)

A
Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

B
Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

C
Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

D
Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Solution

Concept
As per IS 456: 2000, Clause 26.2.1.1,
In limit state method of design, the design bond strength for plain bars in tension are given below in the table:
Grade of Concrete
M20
M25
M30
M35
M40 and above
Design Bond Stress
f_bd N/mm2
1.2
1.4
1.5
1.7
1.9

For deformed bars (e.g. HYSD bars) conforming to IS 1786, these values shall be increased by 60 per cent.
For bars in compression, these values of bond stress for bars in tension shall be increased by 25 per cent.
Development Length
As per IS 456: 2000, clause 26.2.1,
The development length is given by:
\({L_d} = \frac{{\phi \times {f_s}}}{{4 \times {f_{bd}}}}\)
Where
ϕ = nominal diameter of the bar
f_s = stress in the bar at the section considered at design load
τ_bd = design bond stress
Calculations:
The development length is given by:
\({L_d} = \frac{{\phi \times {f_s}}}{{4 \times {f_{bd}}}}\)
a) For deformed bars
f_bd(new) = 1.60 × f_bd
Therefore,
\({L_d} = \frac{{\phi \times {f_s}}}{{4 \times 1.6 \times {f_{bd}}}}\)
\({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)
b) For compression bars
f_bd(new) = 1.25 × f_bd
Therefore
\({L_d} = \frac{{\phi \times {f_s}}}{{4 \times 1.25 \times {f_{bd}}}}\)
\({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)
Question 7
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In the design of a reinforced concrete beam the requirement for bond is not getting satisfied. The economical option to satisfy the requirement for bond is by

A
Bundling of bars

B
Providing smaller diameter bars more in number

C
Providing larger diameter bars less in number

D
Providing smaller diameter bars less in number

Solution

Concept:
Flexural Bond develops due to variation in BM along the reinforcement. (or) Presence of shear force produces flexure bond.
We know Bond stress (τ _bd) is given by
\({\tau _{bd}} = \frac{V}{{\left( {n \times \pi \times \phi } \right) \times L.A}}\)
Where, n is number of bars
Shear force (V ) = dM/dx
ϕ is diameter of the bar
L.A is lever arm
τ_bd should be less than permissible value,
Following are the permissible design bond strength
Grade of concrete
M20
M25
M30
M35
τ_bd(N/mm²)
1.2
1.4
1.5
1.7
As, τ ∝ 1/n
τ ∝ 1/ϕ
If it is greater than (τ_bd)_permissiblethen best economical solution is to reduce the diameter of bar and increase in number.
∴ In case of bond failure most economical way to make structure safe in bond is by providing more number of bars of smaller diameter instead of less number of bars of larger diameter, for the same amount of steel (As, Surface area increases τ decreases which makes the structure safe.)
Question 8
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Reduction coefficient taken for calculating axial load carrying capacity of long column as per working stress method, for the given column dimension is______
(Column dimension = 300 × 400 mm of L_effective = 4500 mm)

A
1.2500

B
1.0156

C
1.1161

D
0.9375

Solution

Concept:
For axial load carrying capacity of column according to working stress method:
\({{\rm{P}}_{{\rm{long}}}} = {{\rm{C}}_{\rm{r}}} \times {{\rm{P}}_{{\rm{short}}}}\)
Cr = Reduction coefficient
\({{\rm{C}}_{\rm{r}}} = 1.25 - \frac{{{{\rm{L}}_{{\rm{effective}}}}}}{{48 \times {\rm{b}}}}\)
Calculation:
\({{\rm{C}}_{\rm{r}}} = 1.25 - \frac{{{{\rm{L}}_{{\rm{effective}}}}}}{{48 \times {\rm{b}}}} = 1.25 - \frac{{4500}}{{48 \times 300}} = 0.9375\)
Question 9
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A rectangular simply supported beam has factored shear force at the centre of the support is 200 KN. The beam has 2 bars of 16 mm dia Fe 415 steel at the centre of the support, on tension zone. M20 grade concrete with τ_bd = 1.2 MPa is used in the beam. Moment of resistance of the beam at support is 60 KN - m. What is the minimum extension of the main steel bars (in mm) with 90° bend to satisfy anchorage bond?

A
230-240

B
2304-2401

C
2307-2406

D
2309-2402

Solution

Concept:
For simply supported beam,
\({\left( {{L_d}} \right)_{actual}} \le \frac{{1.3M}}{V} + {L_o}\)
Where, (L_d)_actual = L_d - Anchorage value
L_o = minimum length of extension
L_d= \(\frac{{\phi {{\rm{\sigma }}_{st}}}}{{4\;{\tau _{bd}}}}\)
Calculation:
Now,
\(\frac{{\phi {{\rm{\sigma }}_{st}}}}{{4\;{\tau _{bd}}}} - AV\; = \;\frac{{1.3M}}{V} + {L_o}\)
\(\Rightarrow \;\frac{{16\; \times \;0.87\; \times \;415}}{{4\; \times \;1.6\; \times \;1.2}} - 8\; \times \;16\; = \;\frac{{1.3\; \times \;60\; \times \;{{10}^6}}}{{200\; \times \;{{10}^3}}} + {L_o}\)
⇒ L_o = 234.18 mm
Question 10
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A short column having cross section 460 × 600 mm (overall depth). It is given that the column is meeting the general requirements of the relevant code for the minimum eccentricity (e_min) as per IS 456 : 2000. If 10 bars of 20 mm diameter is provided as main bar in the column. Calculate the ultimate load carrying capacity of the column (in kN) as per IS 456 : 2000. Use M - 25 concrete and Fe - 415 steel?

A
3601-3603

B
36014-36031

C
36017-36036

D
36019-36032

Solution

Concept
As it is given that, the column is meeting the general requirements of minimum eccentricity
So there is no need to check for minimum eccentricity.
Ultimate load carrying capacity of column (P_U) by Limit state method is given by
P_u = 0.40 f_ck A_c + 0.67 f_y Asc
Where, A_c is Area of concrete = Gross area - Area of steel
A_sc is area of compression steel
Calculation
Given,
number of bars (n) = 10 , dia = 20 mm
M25 Grade of concrete, Fe415 Grade of steel
b = 460 mm , D = 600 mm
\({A_{sc}} = 10 \times \frac{\pi }{4} \times {20^2} = 3141.60\;m{m^2}\)
P_u = 0.40 f_ck A_c + 0.67 f_y Asc
P_u = 0.40 × 25 [A_g - A_sc] + 0.67 × 415 × Asc
A_g= 460 × 600 = 276000 mm²
P_u = 0.40 × 25 [276000 - 3141.60] + 0.67 × 415 × 3141.60
p_u = 3602.10 kN
∴ Ultimate load carrying capacity of column (PU) = 3602.10 kN
Question 11
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At the limit state of collapse, an RC beam is subjected to flexural moment of 200 kN-m, Shear force 20 kN, and torque 9 kN-m. The beam is 300 mm wide and gross depth of 425 mm, with an effective cover of 25 mm. The equivalent nominal shear stress (τ _ve) as calculated by using design code turns out to be lesser than the design shear strength (τ _c) of the concrete.

A
Equivalent shear force is 40 kN

B
Equivalent flexural moment required for longitudinal reinforcement is 212.79 kN

C
Equivalent flexural moment required for longitudinal reinforcement is 200 kN

D
Equivalent shear force is 68 kN

Solution

Concept

Equivalent Shear force is given by

\({V_{eq}} = {V_U} + 1.6 \times \frac{{{T_U}}}{B}\)

Where, T is torque

B is width of the section

V is shear force

Equivalent Moment is given by

\({M_{eq}} = \frac{{{T_U}}}{{1.7}} \times \left( {1 + \frac{D}{b}} \right)\)

Note: If (τ _ve) < (τ _c) then section is designed as follows

Longitudinal Reinforcement is designed for M_Uonly

Shear Reinforcements designed for Nominal shear is provided

\(\frac{{{A_{sv}}}}{{b{S_v}}} > \frac{{0.4}}{{0.87\;{f_y}}}\)

Calculation

Given,

V_U = 20 kN, T_U = 9 kN-m,

B = 300 mm, D = 450 mm

M_U = 200 kN-m

Equivalent nominal shear stress (τ _ve) lesser than the design shear strength (τ _c) of the concrete

(τ _ve) < (τ _c) < (τ _{c, max})

i) Equivalent shear force

\({V_{eq}} = V + 1.6 \times \frac{T}{B}\)

\({V_{eq}} = 20 + \frac{{1.6 \times 9}}{{300 \times {{10}^{ - 3}}}}\) = 68 kN

ii) Equivalent flexural moment

As τ _eq < τ_c section will be designed for M_Uonly.

∴ Equivalent moment = subjected flexural moment = 200 Kn-m
Question 12
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For a RC column fixed at both ends having length of 15 m and cross-section 400 mm × 550 mm. If 4 – 25ɸ bars are used in the column. The load carrying capacity of the column is _____ kN
Take σ_cc = 5 mpa and σ_sc = 195 mpa

A
1090-1100

B
10904-11001

C
10907-11006

D
10909-11002

Solution

Concept
Effective length (l_eff) for both end fixed = 0.65 × L
Slenderness ratio (λ ) > 12 then the column is long given by
\({\rm{λ }} = \frac{{{l_{eff}}}}{{least\;lateral\;Dimension}}\)
In working stress method
Load carrying capacity of long column( P ) = Cr (σcc Ac + σsc ASC)
Where,
C_r is Reduction factor given by
\({C_r} = 1.25 - \frac{{{l_{eff}}}}{{48\;B}}\) 'B' is the Least lateral dimension
A_c is area of concrete
σ _cs is stress in compression steel, σ _ccis stress in concrete
A_g is gross area, A_sc is area of compression steel
Calculation
Given,
L = 15 m, B = 400 mm = 0.4 m
check for slenderness ratio λ
\({\rm{λ }} = \frac{{{l_{eff}}}}{{least\;lateral\;Dimension}}\)
l_eff = 0.65 × L = 0.65 × 15 , l_eff = 9.75 m
least lateral dimension = 0.4 m
\(λ = \frac{{9.75}}{{0.4}} = 24.375 > 12\)
So, it is a long column
P = C_r (σ_cc A_c + σ_sc A_SC)
\({A_{sc}} = 4 × \frac{\pi }{4} × {(25)^2} = 1963.50\;m{m^2}\)
A_c = (400 × 550) – 1963.50, A_c = 218036.5 mm²
\({C_r} = 1.25 - \frac{{{l_{eff}}}}{{48\;B}}\)
\({C_r} = 1.25 - \frac{{9.75}}{{48 × 0.4}}\)
C_r = 0.742
P = 0.742 [5 × 218036.5 + 195 × 1963.50]
P = 1093014.23 N
∴ P = 1093.01 kN

Question 13

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The positive bending moment coefficient at the middle of the end-span of a continuous one way slab is

\(\left( {\frac{{{w_l}}}{{10}} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{10}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{{12}} + \frac{{{w_d}}}{{16}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

Solution

Concept

w_L= Live load

w_d = Dead load

As per Table 12 of IS-456:2000, the bending moment coefficients for dead and live load are given below:

Location	BM Coefficients for dead load	BM Coefficients for Live load	Bending moment
Middle of end span	1/12	1/10	\(\left( {\frac{{{w_l}}}{{10}} + \frac{{{w_d}}}{{12}}} \right){L^2}\)
Interior support of end span	-1/10	-1/9	\(-\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{10}}} \right){L^2}\)
Middle of intermediate span	1/16	1/12	\(\left( {\frac{{{w_l}}}{{12}} + \frac{{{w_d}}}{{16}}} \right){L^2}\)
Interior support of intermediate span	-1/12	-1/9	\(-\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

∴ Option 1 is correct

Question 14
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A prestressed beam of size 250 × 400 mm deep is prestressed by 14 tendons of 6 mm diameter. The cable is located at 150 mm from bottom of the beam. If the initial prestress in the beam is 1250 N/mm². The loss of prestress due to elastic deformation and creep of concrete respectively will be?
Assume:
M-40 Grade concrete
creep coefficient = 1.6
Modulus of elasticity = 2 × 10⁵ MPa.
Neglect the effect due to dead load.

A
54.72 MPa and 87.55 MPa

B
37.13 MPa and 59.42 MPa

C
25.41 MPa and 40.65 MPa

D
42.30 mPa and 55 MPa

Solution

Concept
i) Loss due to elastic shortening = mfc
fc = compressive stress at the level of steel
m = modular ration \(= \frac{{{E_s}}}{{{E_c}}}\)
ii) Loss due to creep of concrete = mϕfc
ϕ = creep coefficient
Calculation
Given,
E_s = 2 × 10⁵ , Grade of concrete M40
Initial prestress = 1250 N/mm²
Number of bars = 14, dia of bar = 6 mm.
Prestressing force (P) = \(14 × \frac{\pi }{4} × {\left( 6 \right)^2} × 1250 = 494.80\;kN.\)
Loss due to elastic shortening = mf_c
f_c = compressive stress at the level of steel
m = modular ration \(= \frac{{{E_s}}}{{{E_c}}}\)
\(m = \frac{{2 × {{10}^5}}}{{5000\sqrt {{F_{ck}}} }} = \frac{{2 × {{10}^5}}}{{5000\sqrt {40} }} = 6.32\)
\({f_c} = \frac{P}{A} + \frac{{Pe}}{I}.y\)
e is eccentricity
At the location of steel tendon, y = e
\({f_c} = \frac{{494.80 × {{10}^3}}}{{250 × 400}} + \frac{{494.80 × {{10}^3} × {{\left( {50} \right)}^2} × 12}}{{250 × {{\left( {400} \right)}^3}}}\)
f_c = 4.948 + 0.9277
f_c = 5.876 MPa
i) loss due to elastic shortening = 6.32 × 5.876
= 37.13 MPa.
Loss due to creep of concrete = mϕf_c
ϕ = creep coefficient = 1.6
ii) loss due to creep of concrete = 6.32 × 1.6 × 5.876
= 59.42 MPa.
Question 15
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Consider a bar of Diameter ‘D’ of length (L) embedded in a large concrete block. If a pull out force P being applied. Let σ_band σ_st be the bond strength (between the bar and concrete) and the tensile strength of the bar respectively. If the block is held in position and material of the block does not fall, which of the following options represents the maximum value of P?

A
Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\)and (π × D × L× σ _st)

B
Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)and (π × D × L× σ_b)

duplicate options foud. English Question 1 options 1,4

C
Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) and (π × D × L× σ _b)

D
Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\) and (π × D × L× σ st)

Solution

Concept
We know,
In an Anchorage bond
Maximum force resisted by bond strength = (π × D × L× σ _b)
Maximum tensile force resisted by bar = \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)
Where,
L is development length
D is dia of bar,
σ_bis bond strength (between the bar and concrete)
σ_st is the tensile strength of the bar
P is pull out Force.
Maximum Pull should be minimum of both resistance for the safety of the bar. So it should not fail in any of the above case.
∴ Maximum value of P = Minimum ( (π × D × L× σ _b) or \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) )
Question 16
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For a reinforced concrete beam with M 20 concrete with Fe 415 steel, the working moment corresponding to ‘balanced failure’ by the Limit Sate Method of Design is A times of the relatable magnitude of the working moment computed by the Working Stress Method. This value of A is nearly

A
1.0

B
1.5

C
2.0

D
2.5

Solution

Concept:
The moment of resistance for balanced section is given as
M = Q Bd²
Where
B is the width of member and d is the effective depth
Q is design coefficient and it depends on the method of design.
As per Limit State Method of Design:
Q = 0.148f_ck for Fe 250
Q = 0.138f_ck for Fe 415
Q = 0.133f_ck for Fe 500
Where f_ck is the characteristic compressive strength.
As per Working Stress Method of Design:
Q = 0.174 σ_cbc for Fe 250
Q = 0.131 σ_cbc for Fe 415
Q = 0.116 σ_cbc for Fe 500
Where σ_cbc is the permissible bending strength for concrete.
Calculation:
For M20: σ_cbc = 7 MPa and f_ck = 20 MPa
Working moment by LSM = \(\frac{{0.138\;{f_{ck}}\;b\;{d^2}}}{{1.5}}\) = 1.84 bd²
Working moment by WSM = 0.131 σ_cbc b d² = 0.131 × 7 bd² = 0.917 bd²
∴ \(\frac{{M\left( {LSM} \right)}}{{M\left( {WSM} \right)}} = A = \frac{{1.84}}{{0.917}} = 2\)
∴ A = 2

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Grade of concrete	M20	M25	M30	M35
τ_bd(N/mm²)	1.2	1.4	1.5	1.7

Steel	X_Umax
Fe250	0.53×d
Fe415	0.48×d
Fe500	0.46×d

RCC Design Test 2

Result

RCC Design Test 2

Score

-

Rank

-

SHARING IS CARING

Question 1

Grade of concrete

Grade of steel

Percentage of steel

All of the above

Solution

Question 2

9-9

94-91

97-96

99-92

Solution

Question 3

25 mm

10 mm

16 mm

20 mm

Solution

Question 4

Singly reinforced beam

Doubly reinforced beam

Beam cannot be designed

Balanced beam

Solution

Question 5

Radius of curvature is 11.16 cm

Effective curvature is 8.96 × 105 mm-1

Effective curvature is 8.96 × 105 mm

Radius of curvature is 11.16 m

Solution

Question 6

Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Solution

Question 7

Bundling of bars

Providing smaller diameter bars more in number

Providing larger diameter bars less in number

Providing smaller diameter bars less in number

Solution

Question 8

1.2500

1.0156

1.1161

0.9375

Solution

Question 9

230-240

2304-2401

2307-2406

2309-2402

Solution

Question 10

3601-3603

36014-36031

36017-36036

36019-36032

Solution

Question 11

Equivalent shear force is 40 kN

Equivalent flexural moment required for longitudinal reinforcement is 212.79 kN

Equivalent flexural moment required for longitudinal reinforcement is 200 kN

Equivalent shear force is 68 kN

Solution

Question 12

1090-1100

10904-11001

10907-11006

10909-11002

Solution

Effective curvature is 8.96 × 10⁵mm^-1

Effective curvature is 8.96 × 10⁵mm

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\)and (π × D × L× σ _st)

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)and (π × D × L× σ_b)

duplicate options foud. English Question 1 options 1,4

Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) and (π × D × L× σ _b)