TGPSC HMWSSB Manager RCC, Steel & Prestressed Design previous year questions with answers
Every RCC, Steel & Prestressed Design question asked in the TGPSC HMWSSB Manager (Engineering) papers on this site, grouped by topic. Answers come from TSPSC's final key; each has a short explanation.
Practise RCC, Steel & Prestressed Design onlineRCC Design
According to IS: 456-2000, in the design of isolated RCC column footing using RCC with an effective depth d, the critical sections to be checked are ______.
- Bending moment at the face of the column, one-way shear at d/2 away from the face of column and two-way shear at d around the column.
- Bending moment at the face of the column, one-way shear at d away from the face of column and two-way shear at d around the column.
- Bending moment at d/2 away from the face of the column, one-way shear at d away from the face of column and two-way shear at d/2 around the column.
- Bending moment at the face of the column, one-way shear at d away from the face of column and two-way shear at d/2 around the column.
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Answer: (4) Bending moment at the face of the column, one-way shear at d away from the face of column and two-way shear at d/2 around the column.
IS 456:2000 (cl. 34.2.3.2 and 34.2.4.1) takes the critical section for bending at the face of the column, for one-way shear at a distance d from the column face, and for two-way (punching) shear on a perimeter at d/2 from the column face.
Minimum amount of high strength deformed bar reinforcement used in solid slabs shall not be less than ______ of the total cross sectional area of the slab according to IS: 456-2000.
- 0.15%
- 0.12%
- 0.20%
- 0.25%
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Answer: (2) 0.12%
IS 456:2000 (cl. 26.5.2.1) requires slab reinforcement of at least 0.12% of the total cross-sectional area when high strength deformed bars or welded wire fabric are used. The value 0.15% applies to mild steel bars.
When the depth of the web of beam is more than ______ mm, side face reinforcement of ______ % of the web area is needed to be provided according to IS:456-2000.
- 600, 0.1%
- 750, 0.15%
- 600, 0.15%
- 750, 0.1%
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Answer: (4) 750, 0.1%
IS 456:2000 (cl. 26.5.1.3) requires side face reinforcement totalling 0.1% of the web area, distributed equally on the two faces, when the depth of the web exceeds 750 mm.
Maximum spacing of vertical shear reinforcement measured along the axis of the RCC beam shall not exceed ______, whichever is lowest.
- 0.75d or 300mm
- 0.75d or 400mm
- 0.50d or 250mm
- 0.5d or 300mm
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Answer: (1) 0.75d or 300mm
IS 456:2000 (cl. 26.5.1.5) limits the spacing of vertical stirrups to 0.75d or 300 mm, whichever is less, where d is the effective depth. For inclined stirrups at 45° the limit is d.
The bearing stress check for column footing in limit state design specifies that the value ______ subjected to a maximum of 2 multiplied by bearing stress ______ shall be more than the compressive stress at the base of the column. A_1 = supporting area for bearing of footing, which in sloped or stepped footing maybe taken as the area of the lower base of the largest frustum of a pyramid or cone contained wholly within the footing and having for its upper base, the area actually loaded and having side slope of one vertical to two horizontal; and, A_2 = loaded area at the column base.
- √(A_1/A_2), 0.30 f_ck
- √(A_1/2A_2), 0.45 f_ck
- √(A_1/A_2), 0.45 f_ck
- √(A_1/A_2), 0.60 f_ck
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Answer: (3) √(A_1/A_2), 0.45 f_ck
IS 456:2000 (cl. 34.4) allows a design bearing stress of 0.45 f_ck on the loaded area, which may be increased by the factor √(A_1/A_2), limited to 2, when the supporting area is larger than the loaded area. This bearing resistance must be at least the compressive stress at the column base; otherwise the excess force has to be carried by reinforcement such as dowels.
The functionality of a wall, retaining wall and a shear wall in order is ______. i) resist predominantly vertical loads ii) resist lateral loads perpendicular to the plane of wall iii) resist lateral loads in the plane of wall
- ii, iii, i
- i, ii, iii
- iii, ii, i
- ii, i, iii
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Answer: (2) i, ii, iii
An ordinary (load-bearing) wall mainly carries vertical gravity loads, a retaining wall resists lateral earth pressure acting perpendicular to its plane, and a shear wall resists lateral wind or earthquake forces acting in its own plane. So the order is i, ii, iii.
Splicing of reinforcement in flexure members is taken-up at a location where bending moment is less than ______ the moment of resistance at that section and not more than ______ of bars are spliced at any particular section.
- 75%, 50%
- 50%, 75%
- 25%, 50%
- 50%, 50%
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Answer: (4) 50%, 50%
IS 456:2000 (cl. 26.2.5.1) requires splices in flexural members to be placed where the bending moment is not more than 50% of the moment of resistance, and not more than half (50%) of the bars may be spliced at any one section.
Steel Design
In a steel tension member, it's maximum strength in case of adequately designed bolted connections is governed by ______.
- The slenderness of the member
- The tensile strength of net sectional area of the member.
- The strength of the bolted connection.
- The section modulus of the member
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Answer: (2) The tensile strength of net sectional area of the member.
When the bolted connection is adequately designed it does not fail first, so the member's strength is limited by rupture of its net cross-section at the bolt holes. Slenderness and section modulus govern compression and bending members, not tensile strength.
The principal reason for adding stiffeners to the web of a steel beam is to ______.
- Increase its moment carrying capacity.
- Reduce the deflection of the beam.
- Increase the stiffness of the web.
- Improve aesthetics.
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Answer: (3) Increase the stiffness of the web.
Web stiffeners stiffen the thin web so that it does not buckle under shear or cripple and buckle under concentrated loads and reactions. Moment capacity and deflection depend mainly on the flanges and the overall section, not on the stiffeners.
In the built-up steel columns, the lacing is provided ______. i) to keep all individual sections together. ii) to take-up lateral shear to an extent of 2.5% of axial force. iii) to increase the bending strength of the section. iv) to increase the compressive strength of the section.
- i, iii
- ii, iii
- iii, iv
- i, ii
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Answer: (4) i, ii
Lacing ties the component sections of a built-up column together so that they act as one unit, and IS 800 requires it to be designed for a transverse (lateral) shear of 2.5% of the axial force in the column. It does not by itself increase the bending or compressive strength of the section.
The list of principal components of a plate girder is ______. i) Top and bottom flange plates and web plates. ii) Horizontal stiffeners, Intermediate and bearing stiffeners. iii) Cleat angles and seat angles. iv) Web splicing and flange splicing.
- i, iii, iv
- i, ii, iii
- i, ii, iv
- ii, iii, iv
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Answer: (3) i, ii, iv
A plate girder is made up of flange plates and a web plate, horizontal, intermediate and bearing stiffeners, and web and flange splices. Cleat angles and seat angles belong to beam-to-column connections, not to the girder itself.
The requirements to be satisfied in the design of a gantry girder are that it has to withstand effects due to ______.
- Moving loads, impact effects and fatigue.
- Moving loads, longitudinal loads, lateral loads, impact effects and fatigue.
- Lateral loads, longitudinal loads, impact effects and fatigue.
- Dynamic loads, longitudinal loads, impact effects and fatigue.
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Answer: (2) Moving loads, longitudinal loads, lateral loads, impact effects and fatigue.
A gantry girder must resist the vertical moving wheel loads of the crane with impact, lateral (surge) loads from the crab, longitudinal loads from braking of the crane, and fatigue from repeated load cycles. Only option (2) lists all of these.
The Lug angle is a member which ______.
- is connected to the main tension member to transfer the tensile force economically to the joint.
- is connected to the main tension member for erection purpose.
- is connected to the main tension member to increase its strength locally.
- is connected to the main tension member to increase its stiffness
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Answer: (1) is connected to the main tension member to transfer the tensile force economically to the joint.
A lug angle is a short angle connected to the outstanding leg of the main member and to the gusset, so that part of the force is passed to the gusset directly. This shortens the end connection and transfers the tensile force to the joint economically.
The expression working out the thickness of slab base is given by ______. If t = thickness of slab base, w = pressure under slab base, σ_bs = permissible bending stress in slab base, a, b = longer and shorter projections of the slab base edge to the column member, Poisson's ratio = 0.25.
- t = ((3w/σ_bs)(a-b/4))^(1/2)
- t = ((3w/σ_bs)(a²-b²/4))^(1/2)
- t = ((3w/σ_bs)(a²-b²))^(1/2)
- t = ((3w/σ_bs)(a-b²/4))^(1/2)
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Answer: (2) t = ((3w/σ_bs)(a²-b²/4))^(1/2)
For a slab base under uniform pressure w, the working-stress formula (IS 800:1984) is t = √[(3w/σ_bs)(a² − b²/4)], where a and b are the greater and lesser projections beyond the column. The b²/4 term allows for two-way bending of the plate with Poisson's ratio 0.25; options (1) and (4) are dimensionally inconsistent and option (3) uses b² instead of b²/4.
Prestressed Concrete
A simply supported beam of span L and uniform flexural rigidity EI is subjected a pre-stressing force of magnitude P, at an eccentricity e below the neutral layer, parallel to the axis of the member. The displacement of the beam at a location L/4 from the support is ______.
- PeL²/4EI
- PeL²/8EI
- PeL²/2EI
- PeL²/EI
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Answer: (2) PeL²/8EI
A straight tendon at eccentricity e gives a uniform moment M = Pe along the span, and a simply supported beam under uniform moment cambers upward by ML²/8EI = PeL²/8EI, the keyed value (this is the mid-span camber). Strictly, at L/4 the expression Mx(L − x)/2EI gives 3PeL²/32EI, which is not among the options. (Official final key; some sources differ.)
The loss stress due to creep in steel in a pre-stress problem is given by the formula ______, where α = creep coefficient, f_c = stress in concrete, E_c = modulus of elasticity of concrete and E_s = modulus of elasticity of steel.
- α(f_c/E_c) E_s
- α(f_c/2E_c) E_s
- α(f_c/E_s) E_c
- 2α(f_c/E_c) E_s
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Answer: (1) α(f_c/E_c) E_s
The elastic strain of concrete at the steel level is f_c/E_c and creep adds α times this strain; the steel shortens by the same amount. So the loss of prestress = α(f_c/E_c)E_s, i.e. α·m·f_c with modular ratio m = E_s/E_c.
The principal reason for adopting pre-stressing cable profiles in flexure members as parabolic is due to the fact that ______.
- They need to resist both bending moments and shear forces in members.
- The strength of pre-stressing cables is maximum in parabolic shapes only.
- The profile of moment caused by self weight of structure is parabolic and to counter this, the cable profile also needs to be parabolic.
- Its a regular practice
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Answer: (3) The profile of moment caused by self weight of structure is parabolic and to counter this, the cable profile also needs to be parabolic.
Self weight and other uniformly distributed loads give a parabolic bending moment diagram, so a parabolic tendon produces a prestressing moment P·e(x) of the same shape that counteracts it at every section (load balancing). The strength of the cable itself does not depend on its profile.