In this post, I will share some important questions which are frequently asked in competitive examinations such as SSC, UPSC and state PSCs.
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How diagonal tension reinforcement is provided in a beam?
- Longitudinal Bars
- Helical reinforcement
- Bent-up bars
- None of the above
View AnswerAnswer: c -
Why minimum shear reinforcement is provided in a beam?
- To resist the bond stress at the support
- To resist shear at the support
- To arrest the longitudinal cracks on side faces due to shrinkage and temperature variation.
- All of the above
View AnswerAnswer: c
The minimum shear reinforcement is provided in a beam so that beam does not fail in shear when the shear strength of concrete gets reduced due to tension-induced into the beam through shrinkage or some restraint. -
The maximum value of permissible shear stress is based on ...
- Diagonal tension failure
- Diagonal compression failure
- Flexural compression failure
- Flexural tension failure
View AnswerAnswer: b
The maximum permissible shear stress is based on diagonal compression failure because if the total area of stirrups and bent-up bars exceed a certain limit, the section might become stronger in diagonal tension as compared to diagonal compression. Hence, a shear compression failure may occur even before the shear reinforcement has yielded and that is an undesirable situation as the nature of the failure is brittle. -
The first crack formed in a deep beam is ...
- Diagonal crack due to tension
- Diagonal crack due to compression
- Shear crack
- Flexural crack
View AnswerAnswer: a -
What will be the maximum spacing between vertical stirrups in an RCC beam when the effective depth is 300 mm?
- 250 mm
- 300 mm
- 275 mm
- 225 mm
View AnswerAnswer: d
The maximum spacing is the minimum of 0.75 times the effective depth of beam and 300 mm. -
An RCC beam of effective depth 1 m and effective span 10 m is supported on columns of size 500 mm X 500 mm. If the beam is loaded with a UDL of 10 MN/m then what is the design shear force?
- 37.5 MN
- 50 MN
- 43 MN
- 47.5 MN
View AnswerAnswer: a
The critical section is 1 m from the face of the support and 1.25 m from the centre of the support.
shear force at support is 50 MN
Design Shear force $=50-10\times (0.25+1)=37.5$ MN -
Which of the following is correct for compression steel?
- ${L}_{d}=\frac{\phi {\sigma}_{s}}{4 {\tau}_{bd}}$
- ${L}_{d}=\frac{\phi {\sigma}_{s}}{5 {\tau}_{bd}}$
- ${L}_{d}=\frac{\phi {\sigma}_{s}}{6.4 {\tau}_{bd}}$
- ${L}_{d}=\frac{\phi {\sigma}_{s}}{8 {\tau}_{bd}}$
View AnswerAnswer: b
In the case of compresstion steel the bond stress is increased by 25%. -
How bond stress gets changed when HYSD bars are used?
- Its value increased by 25%
- Its value decreased by 25%
- Its value increased by 50%
- Its value decreased by 50%
View AnswerAnswer: c -
The term "${M}_{1}/V$" in the expression ${L}_{d}\leq {M}_{1}/V + {L}_{0}$ is increased by ____ when the reaction is compressive.
- 25%
- 30%
- 45%
- 60%
View AnswerAnswer: b
As per IS 456, Cl 26.2.3.3, the term ${M}_{1}/V$ is increased by 30%. -
In above question, what is the value of ${L}_{0}$?
- The effective depth of the member
- 12 times the diameter of the bar
- Smaller of (a) and (b)
- Greater of (a) and (b)
View AnswerAnswer: d
As per IS 456, CL 26.2.3.3, it is greater of the first two options. -
Which of the following statements is incorrect?
- If the bars are larger than 36 mm in diameter, Lap splices are not generally provided.
- Lap length including anchorage value of hooks for the bar in flexural tension shall be ${L}_{d}$ or $12 \phi$, whichever is greater.
- The straight length of the lap shall not be less than $12 \phi$ or 200 mm.
- None of the above
View AnswerAnswer: c
As per IS 456, Cl 26.2.5.1 [c], it is $15 \phi$ or 200 mm. -
If "D" is the diameter of the bar, then what is the anchorage value of the U-hook?
- 12D
- 16D
- 18D
- 20D
View AnswerAnswer: b
As per IS 456, Cl 26.2.2.1 [b-2], it's "16D" -
What is the radius of a U type hook (Mild steel)?
- 4 times the diameter
- 3 times the diameter
- 2 times the diameter
- none of the above
View AnswerAnswer: c
As per IS 2502, Fig 1, (Note), the radius is "KD" and K is 2 for mild steel and 4 for HYSD bar. -
When the splices are welded in compression then the strength of the weld is taken as ...
- 100 % of the design strength of the joined bars
- 80 % of the design strength of the joined bars
- 90 % of the design strength of the joined bars
- 750 % of the design strength of the joined bars
View AnswerAnswer: a
As per IS 456, Cl 26.2.5.2, it is 100 % of the design strength of the joined bars when the splices are in compression.
If the bars are spliced in tension then the strength of weld is 80 % of the design strength of the welded bars and 100 % of design strength of the mechanical connection.
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