Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which set correctly identifies the currently applicable principal Indian Standards for dead, imposed and wind loads relevant to a steel roof truss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 875 (Part 1):2026 is the current Indian Standard for dead loads and unit weights of building, construction and stored materials. It replaced the earlier 1987 edition after the withdrawal date specified by BIS in September 2026. IS 875 (Part 2):1987 remains the applicable standard for imposed loads and was reaffirmed in 2023. IS 875 (Part 3):2015, together with its applicable amendments, governs wind loads. Structural design of the steel truss members themselves is then carried out using IS 800:2007 and its applicable amendments. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Under the usual idealized analysis of a roof truss, which statement is most appropriate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A conventional truss analysis idealizes the joints as pins and the member centrelines as meeting at the joints. Loads are ideally introduced at these panel points. Under these assumptions, an individual member behaves essentially as a two-force member. Its primary internal action is therefore axial tension or axial compression rather than bending. Real connections may introduce secondary effects, but the axial-force model remains the fundamental basis of ordinary roof-truss analysis. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which sequence most accurately represents the gravity-load path in a conventional steel roof-truss system?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Roof sheeting or other roof covering first receives the distributed roof load. Purlins support that covering and span between adjacent roof trusses. The purlin reactions are preferably introduced into the trusses at top-chord panel points. The triangulated truss then develops axial member forces that transfer the actions toward its supports. The supporting columns or walls finally carry these reactions to the foundations. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
Which design approach most reasonably promotes an economical steel roof truss without imposing an arbitrary universal optimum geometry?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An economical truss is not obtained from one universal span-to-rise or panel-size rule. Efficient triangulation allows loads to be carried mainly through axial member action. Reasonable panel dimensions help control member length, slenderness and the number of costly connections. Aligning purlin reactions with panel points also reduces undesirable local bending of the chords. Member sizes should be selected according to their actual force, stability and connection requirements rather than made unnecessarily uniform or heavy. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
A roof truss spans 12 m and adjacent trusses are spaced 4 m apart. If the total permanent roof dead load has already been evaluated as 0.35 kN/m^2 of horizontal plan area, what total dead load is tributary to one truss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The horizontal tributary plan area for one truss is its span multiplied by the spacing of adjacent trusses. This area is 12 × 4 = 48 m^2. The stated equivalent permanent load is 0.35 kN/m^2 of that plan area. Hence the total load is 0.35 × 48 = 16.8 kN. Permanent actions can include the roof covering, purlins, truss self-weight and other permanently attached components as applicable. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
According to IS 875 (Part 2):1987, what uniformly distributed imposed load on plan area applies to a roof having slope not exceeding 10 degrees when access is not provided except for maintenance?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 875 (Part 2) specifies roof imposed loads according to roof geometry and accessibility. For a flat, sloping or curved roof with slope up to and including 10 degrees, where access is not provided except for maintenance, the specified uniformly distributed imposed load is 0.75 kN/m^2 on plan area. Other minimum concentrated or distributed requirements can also apply to members directly supporting roof coverings. The value should not be confused with the higher 1.5 kN/m^2 applicable where access is provided for the relevant roof category. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A sloping roof without general access has a slope of 25 degrees. Using the IS 875 (Part 2):1987 expression w = 0.75 - 0.02(theta - 10) kN/m^2, subject to a minimum of 0.40 kN/m^2, what imposed load applies to the roof membrane sheets or purlins?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The roof slope exceeds 10 degrees, so the specified reduction formula is applicable. The amount of slope beyond 10 degrees is 25 - 10 = 15 degrees. The reduction is therefore 0.02 × 15 = 0.30 kN/m^2. The calculated imposed load is 0.75 - 0.30 = 0.45 kN/m^2, which is above the minimum value of 0.40 kN/m^2. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
Using the same IS 875 (Part 2):1987 roof-imposed-load rule, what value applies to roof membrane sheets or purlins when the roof slope is 30 degrees?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a 30-degree roof, the slope excess above 10 degrees is 20 degrees. The formula initially gives 0.75 - 0.02 × 20 = 0.35 kN/m^2. IS 875 (Part 2) does not permit this roof imposed load to fall below 0.40 kN/m^2. The minimum therefore overrides the calculated 0.35 kN/m^2. Hence the applicable value is 0.40 kN/m^2 and Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
For the 25-degree roof in Question 7, the imposed load on purlins or roofing sheets is 0.45 kN/m^2. Under the IS 875 (Part 2):1987 provision for members supporting purlins on roofs steeper than 10 degrees, what imposed-load intensity may be used for the roof truss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 875 (Part 2) permits trusses, beams and girders supporting roof purlins on roofs with slope greater than 10 degrees to be designed for two-thirds of the imposed load on the purlins or roofing sheets. The purlin-level imposed load in this case is 0.45 kN/m^2. Two-thirds of 0.45 is 0.30 kN/m^2. This reduction applies to the supporting member in the situation described by the code and should not be confused with the load required for the purlin itself. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which expression gives the design wind speed Vz under IS 875 (Part 3):2015?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 875 (Part 3):2015 determines design wind speed at height z from the basic wind speed modified by relevant factors. The expression is Vz = Vb k1 k2 k3 k4. These factors account for matters including risk or probability, terrain and height, topography and the applicable cyclonic-region importance effect. Wind pressure is subsequently obtained from the square of this design wind speed. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
If the design wind speed at roof level is Vz = 40 m/s, what wind pressure pz is obtained from pz = 0.6Vz^2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The wind-pressure equation uses Vz in metres per second and gives pressure in N/m^2. Substitution gives pz = 0.6 × 40^2 = 960 N/m^2. Dividing by 1000 converts this value to 0.96 kN/m^2. This is the basic design wind pressure at the stated height before application of the additional pressure factors used for the relevant structural calculation. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
For pz = 0.96 kN/m^2, take Kd = 0.90, Ka = 0.80 and Kc = 0.90. IS 875 (Part 3):2015 gives pd = Kd Ka Kc pz but also requires pd not to be less than 0.70pz. What design wind pressure governs?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Direct multiplication gives 0.90 × 0.80 × 0.90 × 0.96 = approximately 0.622 kN/m^2. However, the code specifies that the resulting design wind pressure should not be less than 0.70pz. The lower-bound value is 0.70 × 0.96 = 0.672 kN/m^2. Because 0.672 kN/m^2 exceeds the directly calculated 0.622 kN/m^2, the lower bound governs. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
At a roof surface, take external pressure coefficient Cpe = -0.90, internal pressure coefficient Cpi = +0.20 and design wind pressure pd = 0.80 kN/m^2. Using p = (Cpe - Cpi)pd, what net pressure acts on the surface?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The net pressure coefficient is Cpe - Cpi. Here it is -0.90 - 0.20 = -1.10. Multiplication by 0.80 kN/m^2 gives -0.88 kN/m^2. With the stated sign convention, the negative result represents net suction or uplift on the roof surface. Such uplift can materially change the force pattern in a roof truss compared with downward gravity loading. Therefore Option D is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a conventional simply supported pitched roof truss under symmetric downward gravity loading, which member-force pattern is generally expected?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Downward gravity loading causes a simply supported roof truss to behave globally in a manner analogous to a simply supported beam. The upper chord forms the principal compression chord. The lower chord generally acts as the tension tie. Web members transmit shear between these chords and may be either in tension or compression depending on the truss configuration and load pattern. The precise member forces must still be obtained from structural analysis rather than assumed solely from appearance. Therefore Option A is correct.
వివరణ:
Question 15
ప్రశ్న 15
Why must roof-truss members that are tension members under gravity loading sometimes also be checked for compression?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Wind pressure on a lightweight roof can create substantial uplift. The resulting joint loads can be opposite in direction to the dominant gravity loads. Consequently some members that act as ties under dead and imposed loading can become compression members in a wind-load combination. Compression introduces buckling and slenderness requirements that do not arise from tensile strength alone. The design should therefore consider force reversal whenever the governing load combinations can produce it. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
Roof trusses are spaced 4 m apart and have top-chord panel points at 2 m horizontal intervals. For a roof of slope not exceeding 10 degrees with no access except maintenance, take the roof imposed load as 0.75 kN/m^2 on plan area. What vertical imposed load is transferred to a typical interior top-chord panel point from its tributary roof area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A typical interior panel point receives load from a horizontal tributary width equal to one full panel length. The tributary plan area is therefore 2 m × 4 m = 8 m^2. Multiplication by the stated imposed-load intensity gives 0.75 × 8 = 6 kN. Applying the equivalent concentrated force at the panel point preserves the axial-member idealization of the truss. End panel points would have a different tributary width where only half a panel contributes. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
What is the principal structural consequence if a significant purlin reaction is applied to a top-chord member between two roof-truss panel points rather than at a joint?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Ideal truss action assumes that external loads are applied at the joints. A load acting along a chord between joints causes transverse bending of that chord segment. The chord then carries combined axial force and bending rather than behaving as a pure two-force member. This can require additional member and connection checks. Placing purlins at or close to top-chord panel points is therefore structurally advantageous where the framing arrangement permits it. Therefore Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
At the apex of a symmetrical triangular roof truss, two identical rafters meet and each is inclined at 30 degrees to the horizontal. A vertical downward joint load of 100 kN acts at the apex, and no other member meets that joint. What axial force is developed in each rafter?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
By symmetry, the two rafter forces have equal magnitude. Their horizontal components cancel at the apex. The two upward vertical components must together balance the 100 kN downward load. Therefore 2F sin30 degrees = 100 kN, giving F = 100 kN. The forces must be compressive because their actions at the apex need upward components to resist the applied downward load. Therefore Option D is correct.
వివరణ:
Question 19
ప్రశ్న 19
For the same three-member triangular truss in the previous question, what axial force is developed in the horizontal bottom tie?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Each rafter carries 100 kN compression and is inclined at 30 degrees to the horizontal. Its horizontal force component is 100 cos30 degrees = approximately 86.6 kN. At an eave joint this horizontal component must be balanced by the bottom chord under the ordinary pin-roller support idealization of the complete truss. The bottom chord therefore develops approximately 86.6 kN of tensile force. This illustrates the tie action that prevents the two rafters from spreading apart. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
According to IS 800:2007, what effective length may generally be used for an in-plane compression member of a bolted, riveted or welded truss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800 gives a specific effective-length provision for compression members in trusses and braced frames. For in-plane buckling, the effective length can be taken between 0.7 and 1.0 times the distance between centres of connections. The selected value depends on the degree of end restraint actually available to the member. Greater rotational restraint can justify a value toward the lower end of this range. The provision does not mean that every truss compression member automatically has an effective-length factor of exactly 0.7. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
A roof-truss compression member is 3.0 m long between connection centres. For the assessed in-plane end restraint, take KL = 0.8L. If its relevant radius of gyration is 30 mm, what is its effective slenderness ratio?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The effective length is 0.8 × 3000 = 2400 mm. Effective slenderness is calculated as KL/r. Substitution gives 2400/30 = 80. This slenderness is then used with the appropriate buckling class and material properties to obtain the compression-member design strength. The calculation must be performed about the relevant possible buckling directions, not just whichever axis gives the more favourable result. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
What effective-length principle does IS 800:2007 specify for buckling of a truss compression member perpendicular to the plane of the truss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A truss compression member may buckle either in the plane of the truss or out of that plane. IS 800 gives a distinct rule for the perpendicular-to-truss-plane buckling mode. The effective length is taken as the distance between the relevant centres of intersection. Actual lateral restraint and the structural path that supplies it therefore matter greatly for a compression chord. A purlin or bracing member should not be assumed to provide effective restraint unless its connection and supporting system can actually develop the necessary restraint. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Which pair of maximum effective slenderness ratios is correctly associated with IS 800:2007 roof-truss applications?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800 limits the effective slenderness of a member carrying compressive force from dead and imposed loads to 180. This controls excessive flexibility as well as practical stability concerns. A member normally acting as a tie in a roof truss or bracing system, and not considered effective when subjected to possible reversal from wind or earthquake, is permitted a larger limiting value of 350. The two limits apply to different structural roles and should not be interchanged. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement correctly describes the design of a bolted angle tension member connected to a roof-truss gusset plate?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 800 requires an axial tension member to be checked against the relevant tensile failure modes. These include gross-section yielding, rupture of the critical effective net section and block shear at the connection. For an angle connected through one leg, shear-lag effects can also influence the net-section rupture strength. The bolts or welds and the gusset plate must themselves be capable of transferring the required factored force. Thus satisfactory member strength alone does not make an inadequate connection acceptable. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
A roof-truss compression member has effective sectional area Ae = 1800 mm^2 and, after the applicable effective-length and buckling-class checks, design compressive stress fcd = 140 MPa. If its factored axial compression is 230 kN, which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The design axial compression strength is Pd = Ae fcd. Substitution gives 1800 × 140 = 252000 N, which is 252 kN. The applied factored compression of 230 kN is therefore below the stated design strength. On this axial-strength check alone, the member is adequate. Final selection must still satisfy the applicable slenderness, local-section, connection and force-reversal requirements of the roof-truss design. Therefore Option C is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Dead loads: IS 875 (Part 1):2026; imposed loads: IS 875 (Part 2):1987; wind loads: IS 875 (Part 3):2015 సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. Members are treated primarily as two-force members carrying axial tension or compression when loads are applied at ideal pin joints సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. Roof covering → purlins → truss panel points/top chord → truss members → supports సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. Use sensible triangulation and panel dimensions, keep member slenderness and connection eccentricity under control, and coordinate purlin locations with panel points సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 16.8 kN సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 0.75 kN/m^2 సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. 0.45 kN/m^2 సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 0.40 kN/m^2 సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 0.300 kN/m^2 సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. Vz = Vb k1 k2 k3 k4 సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: C. 0.96 kN/m^2 సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: B. 0.672 kN/m^2 సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: D. -0.88 kN/m^2 సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. Top chord predominantly in compression and bottom chord predominantly in tension సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: A. Wind uplift can reverse the truss force pattern and change the sign of force in some members సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: B. 6 kN సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: A. The top-chord segment can develop local bending in addition to its truss axial force సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: D. 100 kN compression సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: C. 86.6 kN tension సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. Between 0.7 and 1.0 times the distance between centres of connections, depending on the degree of end restraint సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: B. 80 సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. KL is taken as the distance between the centres of intersection relevant to that out-of-plane buckling mode సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. Compression member carrying dead and imposed loads: 180; member normally acting as a roof-truss tie but not considered effective when wind or earthquake causes possible stress reversal: 350 సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. Its design tension strength is governed by the least applicable strength from gross-section yielding, critical net-section rupture and block shear, while the gusset connection must also safely transfer the factored member force సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: C. Its design compression strength is 252 kN, so the stated axial-strength check is satisfied సమాధానం: C.