Test 10 - Torsion in RCC Members

Question 1 ప్రశ్న 1

Which statement correctly reflects the IS 456 treatment of torsion in reinforced-concrete structures?

A Where torsion is required to maintain equilibrium, the member shall be designed for torsion; in an indeterminate structure, torsion caused by redundant restraint may in appropriate cases be released if torsional stiffness is neglected in determining internal forces. B Every calculated torsional moment must always be resisted regardless of whether it results only from a redundant restraint that can be released. C Torsion may always be ignored whenever shear reinforcement is present. D Only statically determinate members require torsion design.

Question 2 ప్రశ్న 2

Which description best represents the structural behaviour and reinforcement concept of an RCC beam subjected to significant torsion?

A Torsion can produce inclined cracking around the member, and effective resistance requires a combination of closed transverse reinforcement and appropriately distributed longitudinal reinforcement. B Torsion produces only vertical flexural cracks and requires longitudinal reinforcement on one face only. C Torsion is resisted exclusively by the tensile strength of uncracked concrete. D Open stirrups alone provide complete torsional resistance without longitudinal reinforcement.

Question 3 ప్రశ్న 3

How does the IS 456 Limit State approach account for combined bending, shear and torsion in an RCC beam?

A Longitudinal reinforcement is designed using equivalent bending actions, while transverse reinforcement is designed using the combined effects of shear and torsion. B Torsion is designed entirely independently of both bending and shear. C The actual bending moment is ignored whenever torsion is present. D Only an equivalent axial force is calculated.

Question 4 ప్రశ్న 4

For torsion design near the support of an RCC beam, what provision applies to sections located at a distance less than d from the face of the support?

A They may be designed for the same torsional moment as that calculated at a distance d from the face of the support. B The torsional moment must always be taken as zero. C The torsional moment must always be doubled. D They are designed only for bending and no torsion check is permitted.

Question 5 ప్రశ్న 5

For a rectangular RCC beam subjected to factored shear Vu and factored torsional moment Tu, which expression gives the IS 456 equivalent shear Ve?

A Ve = Vu + 1.6Tu/b B Ve = Vu + Tu/D C Ve = Vu + 1.7Tu/b D Ve = 1.6Vu + Tu/b

Question 6 ప్రశ్న 6

A rectangular beam has b = 300 mm, d = 550 mm and D = 600 mm and is subjected to Vu = 120 kN, Tu = 30 kN·m and Mu = 150 kN·m. What is the equivalent shear Ve?

A 200 kN B 240 kN C 280 kN D 320 kN

Question 7 ప్రశ్న 7

For the beam in the previous question, take M25 concrete and assume the relevant Table 19 value is τc = 0.64 N/mm². Which statement correctly describes the equivalent shear check?

A τve is about 1.70 N/mm², which exceeds τc but is below τc,max = 3.10 N/mm², so the required longitudinal and transverse reinforcement provisions for combined actions apply. B τve is about 0.64 N/mm², so torsion has no influence. C τve is about 3.50 N/mm² and the section must necessarily be abandoned. D τve is about 0.17 N/mm² and no reinforcement is needed.

Question 8 ప్రశ్న 8

An M25 beam has b = 300 mm, d = 500 mm, Vu = 30 kN and Tu = 5 kN·m. If the applicable concrete design shear strength is τc = 0.64 N/mm², what follows from the equivalent shear check under the current amended IS 456 provisions?

A τve is about 0.378 N/mm²; minimum shear reinforcement is required and longitudinal reinforcement shall also be provided as required by the torsion provisions. B τve is about 1.20 N/mm², so the section exceeds τc,max. C τve is zero because the torsional moment is relatively small. D Since τve is below τc, neither longitudinal nor transverse reinforcement is required.

Question 9 ప్రశ్న 9

An M25 beam has b = 300 mm, d = 500 mm, Vu = 200 kN and Tu = 55 kN·m. What is the correct conclusion from the equivalent shear check?

A Ve is about 493.3 kN and τve is about 3.29 N/mm², which exceeds τc,max = 3.10 N/mm², so the section should be redesigned. B Ve is about 255 kN and minimum reinforcement alone is sufficient. C τve is about 2.20 N/mm² and no torsion reinforcement is needed. D Any value of τve can be accepted if sufficiently large closed stirrups are provided.

Question 10 ప్రశ్న 10

A T-beam has effective flange width bf = 1000 mm, web width bw = 250 mm, d = 550 mm, Vu = 120 kN and Tu = 25 kN·m. Using the IS 456 torsion procedure for a flanged beam, which pair is correct?

A Ve = 280 kN and τve ≈ 2.04 N/mm² using bw = 250 mm B Ve = 160 kN and τve ≈ 0.29 N/mm² using bf = 1000 mm C Ve = 280 kN and τve ≈ 0.51 N/mm² using bf = 1000 mm D Ve = 120 kN and τve ≈ 0.87 N/mm² using bw = 250 mm

Question 11 ప్రశ్న 11

Which expression gives the torsional equivalent moment Mt used in designing longitudinal reinforcement under IS 456?

A Mt = Tu(1 + D/b)/1.7 B Mt = 1.6Tu/b C Mt = Tu(1 - D/b)/1.7 D Mt = VuD/1.7

Question 12 ప్రశ్న 12

For the rectangular beam with b = 300 mm, D = 600 mm and Tu = 30 kN·m, what is Mt approximately?

A 30.0 kN·m B 42.4 kN·m C 52.9 kN·m D 90.0 kN·m

Question 13 ప్రశ్న 13

The same beam is subjected to Mu = 150 kN·m in addition to Tu = 30 kN·m. What equivalent bending moment Me1 should be used for the principal longitudinal reinforcement design?

A 97.1 kN·m B 150.0 kN·m C 202.9 kN·m D 240.0 kN·m

Question 14 ప్రశ్న 14

A beam has b = 300 mm, D = 600 mm, Mu = 40 kN·m and Tu = 30 kN·m. Since Mt is about 52.94 kN·m, what additional longitudinal design requirement arises?

A Because Mt exceeds Mu, longitudinal reinforcement is also required on the flexural compression face for Me2 = Mt - Mu ≈ 12.94 kN·m acting in the opposite sense. B No longitudinal reinforcement is required because Mu is less than Mt. C The bottom reinforcement is designed only for Mu = 40 kN·m. D Me2 is Mu + Mt = 92.94 kN·m and acts in the same sense as Me1.

Question 15 ప్రశ్న 15

An M25 Fe415 rectangular beam has b = 300 mm, d = 500 mm, D = 600 mm, Mu = 250 kN·m and Tu = 40 kN·m. Mt is about 70.59 kN·m, giving Me1 ≈ 320.59 kN·m, while the singly reinforced limiting moment is about 258.7 kN·m. What is the correct design interpretation?

A The equivalent bending demand exceeds the singly reinforced limiting capacity, so the section dimensions should be increased or an appropriate doubly reinforced design should be adopted. B The beam can remain singly reinforced merely by adding unlimited tension steel. C Torsion may be ignored because Mu alone is below Me1. D The limiting moment applies only to Working Stress design.

Question 16 ప్రశ్న 16

For the M25 Fe415 beam with b = 300 mm, d = 550 mm and Me1 = 202.94 kN·m, what tensile-steel area is required approximately if the equivalent flexural design is singly reinforced and the steel develops 0.87fy?

A 825 mm² B 1000 mm² C 1159 mm² D 1795 mm²

Question 17 ప్రశ్న 17

When torsion reinforcement is required in an RCC beam, which reinforcement arrangement is consistent with IS 456?

A Longitudinal reinforcement together with two-legged closed hoops enclosing the corner longitudinal bars B Open U-stirrups alone with no longitudinal corner reinforcement C Only additional bottom flexural bars with no transverse reinforcement D Only compression reinforcement at the neutral axis

Question 18 ప్రశ్న 18

For combined factored torsion Tu and shear Vu, which expression gives the required area Asv of a two-legged closed hoop at spacing sv under the IS 456 torsion provision?

A Asv = Tu sv/[b1d1(0.87fy)] + Vu sv/[2.5d1(0.87fy)] B Asv = Vu sv/[bd(0.87fy)] only C Asv = Tu sv/[1.6b(0.87fy)] D Asv = 0.36fckb1d1/(fy sv)

Question 19 ప్రశ్న 19

A beam has Tu = 30 kN·m, Vu = 120 kN, b1 = 240 mm, d1 = 500 mm and Fe415 closed hoops at spacing sv = 150 mm. Using the IS 456 combined torsion-and-shear expression, what Asv is required approximately?

A 100.5 mm² B 131.7 mm² C 143.7 mm² D 201.1 mm²

Question 20 ప్రశ్న 20

For the beam in the previous question, b = 300 mm, d = 550 mm, Ve = 280 kN and τc = 0.64 N/mm². At sv = 150 mm, the lower-bound transverse-reinforcement requirement based on (τve - τc)bsv/(0.87fy) is approximately 131.7 mm². Which requirement governs?

A The combined torsion-and-shear expression governs with Asv ≈ 143.7 mm². B The lower-bound requirement governs with Asv ≈ 131.7 mm². C Both requirements may be ignored because τve exceeds τc. D Only minimum longitudinal reinforcement governs.

Question 21 ప్రశ్న 21

For the same beam, two-legged 10 mm diameter Fe415 closed hoops provide Asv ≈ 157.1 mm². The combined torsion-and-shear strength equation permits a maximum spacing of about 163.9 mm, the lower-bound equation permits about 178.9 mm, and the torsion detailing limit is 185 mm. What spacing controls?

A Approximately 164 mm B Approximately 179 mm C 185 mm D 300 mm

Question 22 ప్రశ్న 22

For torsion reinforcement, what is the IS 456 spacing limitation for rectangular closed stirrups in terms of their short and long dimensions x1 and y1?

A The spacing shall not exceed the least of x1, (x1 + y1)/4 and 300 mm. B The spacing shall not exceed the greater of x1 and y1. C The spacing shall not exceed (x1 + y1)/2 with no absolute limit. D The spacing is governed only by 0.75d.

Question 23 ప్రశ్న 23

A rectangular closed torsion stirrup has x1 = 240 mm and y1 = 500 mm. Ignoring any smaller spacing required by strength, what is the maximum spacing permitted by the torsion distribution rule?

A 150 mm B 185 mm C 240 mm D 300 mm

Question 24 ప్రశ్న 24

Which statement correctly describes the distribution of longitudinal reinforcement in a member designed for torsion?

A Longitudinal reinforcement should be placed as close as practicable to the corners, with at least one longitudinal bar at each corner of the ties; when a cross-sectional dimension exceeds 450 mm, additional longitudinal bars are required to satisfy the applicable minimum reinforcement and spacing provisions. B All longitudinal torsion reinforcement should be concentrated at the centroid of the section. C Only two longitudinal bars are permitted regardless of member dimensions. D Corner bars are unnecessary when closed stirrups are used.

Question 25 ప్రశ్న 25

Consider the following statements about the IS 456 Limit State design of members subjected to combined bending, shear and torsion:
1. Equivalent shear is Ve = Vu + 1.6Tu/b.
2. Longitudinal reinforcement is designed for Me1 = Mu + Mt, where Mt = Tu(1 + D/b)/1.7.
3. If Mt exceeds Mu, an additional opposite-sense moment Me2 = Mt - Mu has to be considered for reinforcement on the other face.
4. Rectangular torsion stirrups may be open at one side provided τve is less than τc,max.

Which statements are correct?

A 1 and 4 only B 1, 2 and 3 only C 2, 3 and 4 only D 1, 2, 3 and 4

Answer Key || సమాధానాల పట్టిక

Related Quizzes || సంబంధిత క్విజ్‌లు

Civil HMWSSB, Unit 3 - Cement Concrete and Steel Structures Quiz

TGPSC HMWSSB Quiz

Answer Key సమాధానాల పట్టిక

Related Quizzes సంబంధిత క్విజ్‌లు