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Question 1
ప్రశ్న 1
Which statement correctly describes the current Indian-code context for welded structural-steel connections?
Explanation:
IS 800:2007 contains specific provisions for welded structural-steel connections. Its welding clause requires welds and welding to conform to IS 816 and IS 9595, as appropriate. IS 800 provides Limit State design-strength and detailing rules for structural welds. The published code therefore directly governs many of the structural calculations used in welded connections. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which set consists entirely of weld types recognized in IS 800:2007?
Explanation:
IS 800 recognizes fillet, butt, slot, plug and compound welds for structural connections. A fillet weld is widely used where connected faces meet approximately at right angles. Butt welds join members along their abutting edges and can provide complete or incomplete penetration. Slot and plug welds are specialized forms used in appropriate connection arrangements. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For an ordinary fillet weld, what dimension is taken as the weld size under IS 800:2007?
Explanation:
For a normal fillet weld, IS 800 defines the size by the minimum leg dimension of the weld. The effective throat is a separate quantity used in strength calculations. For a normal 90-degree fillet, the effective throat is approximately 0.7 times the weld size. Deep-penetration welds have additional provisions for recognizing penetration beyond the root run. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A normal 8 mm fillet weld joins two surfaces at 90 degrees. Using the IS 800 value K = 0.70, what effective throat thickness is used for stress calculations?
Explanation:
The effective throat thickness of a fillet weld is K times the specified weld size. For fusion faces at 90 degrees, IS 800 gives K = 0.70. Thus the effective throat is 0.70 × 8 = 5.6 mm. Strength is evaluated on the effective throat area rather than simply on the nominal leg area. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A fillet weld of size 8 mm joins fusion faces at an angle of 110 degrees. If IS 800 Table 22 gives K = 0.55 for the applicable angle range, what effective throat thickness should be used?
Explanation:
The effective throat for inclined fusion faces is obtained as K times the fillet size. An angle of 110 degrees lies in the stated range for which K = 0.55. Hence the throat thickness is 0.55 × 8 = 4.4 mm. The calculation shows that the usual 0.70 factor should not automatically be used for every included angle. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
Two steel parts are to be connected by a single-run fillet weld, and the thicker part is 18 mm thick. What minimum weld size is specified by IS 800:2007 Table 21 in the absence of preheating considerations requiring otherwise?
Explanation:
IS 800 relates the minimum fillet-weld size to the thickness of the thicker connected part. For a thicker part over 10 mm and up to and including 20 mm, the minimum size is 5 mm. The purpose is primarily to reduce the risk of cracking caused by rapid cooling of an undersized weld. The code also states that fillet weld size shall not generally be less than 3 mm. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A fillet weld is applied along the square edge of a 10 mm thick plate. According to the usual IS 800 detailing recommendation, what weld size should generally not be exceeded if the plate edge is to project beyond the weld?
Explanation:
For a fillet weld along the square edge of a part, IS 800 recommends that the specified size generally be at least 1.5 mm less than the edge thickness. This helps avoid washing down of the exposed arris. For a 10 mm plate, the corresponding maximum under this usual recommendation is 10 - 1.5 = 8.5 mm. Other minimum-size and fabrication requirements must also be satisfied. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
A continuous 6 mm fillet weld has an actual overall length of 162 mm. Using the usual IS 800 detailing relation that the actual weld length is the effective length plus twice the weld size, what effective length corresponds to this weld?
Explanation:
The effective length is the portion of weld having the specified full size and required throat thickness. In practice, IS 800 notes that the actual length is made equal to the required effective length plus twice the weld size. Thus the effective length is 162 - 2(6) = 150 mm. The end portions account for the regions over which the full weld profile is developed. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
Under IS 800:2007 Clause 10.5.4.1, a normal continuous fillet weld has size 6 mm. What is the minimum actual weld length under the provision that the actual length shall not be less than four times the weld size?
Explanation:
Clause 10.5.4.1 distinguishes the effective full-size portion of a fillet weld from its actual overall length. In practice, the actual length is taken as the effective length shown on the drawing plus twice the weld size. The same clause states that the actual weld length is not to be less than four times the weld size. For s = 6 mm, this minimum actual length is 4 × 6 = 24 mm. This should not be confused with the separate intermittent-weld rule requiring an effective segment length of at least 4s and at least 40 mm. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
A 6 mm fillet weld terminates at the corner of a connected part. Where an end return is practicable, what minimum return length does IS 800 recommend?
Explanation:
IS 800 recommends returning a fillet weld continuously around the corner where it terminates at the end or side of a part. The return should extend for a distance of at least twice the weld size unless this is impractical. For a 6 mm fillet, the minimum recommended return is therefore 12 mm. End returns are particularly important at tension ends of components subjected to bending. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
Two plates form a welded lap joint, and the thinner plate is 12 mm thick. What minimum lap length is required by the IS 800 rule of four times the thinner-part thickness or 40 mm, whichever is greater?
Explanation:
The minimum lap is the greater of four times the thickness of the thinner connected part and 40 mm. Four times 12 mm is 48 mm. Since 48 mm exceeds 40 mm, the governing minimum lap is 48 mm. This requirement assists development of a practical and effective welded lap connection. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a shop fillet weld, the smaller ultimate tensile strength of the weld metal and parent metal is 410 MPa. Using gamma_mw = 1.25, what design weld strength fwd = fu/(sqrt(3) gamma_mw) is obtained?
Explanation:
The design strength of a shop fillet weld is based on the smaller ultimate stress of the weld and parent material. The nominal weld strength is fu divided by the square root of three. It is then divided by the applicable partial safety factor gamma_mw. Thus fwd = 410/[sqrt(3) × 1.25] = approximately 189.37 MPa. Therefore Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
For the same fu = 410 MPa weld made at site during erection, use gamma_mw = 1.50. What design weld strength is obtained?
Explanation:
IS 800 uses a larger partial safety factor for site welds than for shop welds. For a site weld made during erection, gamma_mw is 1.50. Hence fwd = 410/[sqrt(3) × 1.50] = approximately 157.81 MPa. The resulting design strength is lower than that of the corresponding shop weld. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A shop-welded connection has two parallel 6 mm fillet welds, each with effective length 150 mm. Take K = 0.70 and fwd = 189.37 MPa. What total design strength of the two welds is obtained for direct shear?
Explanation:
The effective throat of each 6 mm weld is 0.70 × 6 = 4.2 mm. The total effective throat area is 2 × 150 × 4.2 = 1260 mm². Multiplying by 189.37 N/mm² gives approximately 238606 N. Thus the total design strength is approximately 238.6 kN. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A factored axial force of 200 kN is to be transferred by two identical parallel 6 mm shop fillet welds. Take K = 0.70 and fwd = 189.37 MPa. What minimum effective length is required for each weld, ignoring long-joint reduction?
Explanation:
Each weld has an effective throat of 0.70 × 6 = 4.2 mm. If each of the two welds has effective length L, their total design resistance is 2L × 4.2 × 189.37 N. Setting this equal to 200000 N gives L = approximately 125.73 mm. A practical design would therefore require at least about 126 mm effective length per weld before other detailing checks. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Two plates of thickness 16 mm and 12 mm are connected by a complete-penetration butt weld. What effective throat thickness is used according to IS 800:2007?
Explanation:
For a complete-penetration butt weld, the effective throat thickness is taken as the thickness of the thinner part joined. The thinner plate in this connection is 12 mm thick. Therefore the effective throat is 12 mm. Reinforcement or excess weld metal above the required profile is not added to this effective throat value. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
How is a complete-penetration butt weld treated for design strength under IS 800:2007?
Explanation:
IS 800 treats butt welds as parent metal having a thickness equal to their effective throat thickness. The stresses in the butt weld must not exceed those permitted in the parent material. A complete-penetration butt weld can therefore provide an efficient force-transfer path when properly detailed and executed. Its design treatment differs from the throat-shear strength expression normally used for fillet welds. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
For an intermittent 6 mm fillet weld, what minimum effective length must each weld segment have under IS 800:2007?
Explanation:
IS 800 specifies that an intermittent fillet-weld segment shall have an effective length of at least four times the weld size. It also specifies an absolute minimum effective length of 40 mm. Four times a 6 mm weld is only 24 mm. The 40 mm absolute minimum therefore governs in this case. Therefore Option D is correct.
వివరణ:
Question 19
ప్రశ్న 19
Intermittent fillet welds connect 10 mm thick plates. What is the maximum permitted clear spacing between effective weld lengths for a tension joint, based on the IS 800 limit of 16 times the thinner-part thickness or 200 mm, whichever is less?
Explanation:
For a tension joint, the clear spacing limit is 16 times the thickness of the thinner connected part. For a 10 mm plate, this gives 16 × 10 = 160 mm. The code also imposes an absolute maximum of 200 mm. Since 160 mm is smaller than 200 mm, the governing clear spacing is 160 mm. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
For the same 10 mm thick plates, what maximum clear spacing applies to intermittent fillet welds in a compression joint when the IS 800 limit is 12 times thickness or 200 mm, whichever is less?
Explanation:
For compression joints, IS 800 uses a limit of 12 times the thinner-part thickness. Twelve times 10 mm is 120 mm. This is below the absolute maximum spacing of 200 mm. The permitted clear spacing is therefore 120 mm. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
A side weld transferring shear has an effective throat thickness of 4.2 mm. At what joint length does the IS 800 long-joint threshold of 150 times the throat thickness occur?
Explanation:
The long-joint provision becomes relevant when the applicable welded-joint length exceeds 150 times the throat thickness. Here the effective throat thickness is 4.2 mm. Thus 150 × 4.2 = 630 mm. A joint longer than this threshold requires the applicable long-joint reduction to be considered where the clause applies. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
A fillet weld is subjected simultaneously to a normal stress fa = 150 MPa and shear stress q = 50 MPa. Using the IS 800 expression fe = sqrt(fa^2 + 3q^2), what equivalent stress is obtained?
Explanation:
The equivalent weld stress is calculated using the specified interaction expression. Substitution gives fe = sqrt(150² + 3 × 50²). The quantity under the square root is 22500 + 7500 = 30000 MPa². Its square root is approximately 173.21 MPa. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
For an in-plane eccentrically loaded weld group, which statement best describes the usual elastic weld-group analysis?
Explanation:
An eccentric in-plane load can be transferred to the centroid of the weld group together with a moment equal to the load times its eccentricity. The direct shear component is distributed over the total effective weld length in the elementary elastic line-weld model. The moment produces a secondary tangential shear proportional to the radial distance from the weld-group centroid. The direct and secondary components are then combined vectorially to locate the most highly stressed point of the weld group. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
A weld group consists of two vertical weld lines, each 100 mm long, located at x = ±50 mm from the group centroid and extending from y = -50 mm to y = +50 mm. A downward load of 100 kN acts 100 mm to the right of the centroid. Using qd = P/Ltotal and qm = Mr/J, with J = 666666.7 mm^3 for the weld-line group, what is the maximum resultant weld force per unit length at a corner?
Explanation:
The total weld length is 200 mm, so the direct downward shear is 100/200 = 0.50 kN/mm. The eccentric moment is 100 × 100 = 10000 kN-mm. At a corner, r = sqrt(50² + 50²) = 70.71 mm, giving secondary shear qm = 10000 × 70.71/666666.7 = approximately 1.061 kN/mm. Its horizontal and vertical components are each approximately 0.75 kN/mm, and at the critical left-hand corners the vertical secondary component acts downward with the direct shear. The resultant is therefore sqrt(0.75² + 1.25²) = approximately 1.458 kN/mm, so Option D is correct.
వివరణ:
Question 25
ప్రశ్న 25
For the eccentric weld group in the previous question, suppose each weld is a 6 mm shop fillet weld with K = 0.70 and fwd = 189.37 MPa. The design resistance per unit effective weld length is t_t fwd. Which conclusion is correct when the maximum demand is 1.458 kN/mm?
Explanation:
The effective throat of a 6 mm normal fillet is 0.70 × 6 = 4.2 mm. Its design resistance per unit effective length is 4.2 × 189.37 = 795.35 N/mm. This is approximately 0.795 kN/mm. The maximum elastic demand from the eccentric weld-group analysis is 1.458 kN/mm, which is substantially higher. The stated 6 mm weld group is therefore inadequate, so Option C is correct.
వివరణ:
Question 26
ప్రశ్న 26
Which statement most accurately compares welded connections with conventional riveted or bearing-type bolted connections?
Explanation:
Welding can connect structural components without drilling the holes required for ordinary riveted or bearing-type bolted joints. It can also provide compact and continuous connection layouts with efficient transfer of force. However, welding introduces heat, which can cause residual stresses and distortion, and workmanship strongly affects weld quality. Fatigue-sensitive welded details require particular care because local geometry and weld imperfections can influence crack initiation. Therefore Option A is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. IS 800:2007 contains structural weld design provisions and requires welds and welding to conform to IS 816 and IS 9595, as appropriate సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. Fillet, butt, slot, plug and compound welds సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: B. The minimum weld leg size సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: B. 5.6 mm సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: B. 4.4 mm సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. 5 mm సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. 8.5 mm సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: B. 150 mm సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: C. 24 mm సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: C. 12 mm సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 48 mm సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: B. 189.4 MPa సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: B. 157.8 MPa సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 238.6 kN సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 126 mm సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. 12 mm సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: B. It is treated as parent metal with thickness equal to the effective throat, and stresses are limited by the corresponding parent-metal design provisions సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. 40 mm సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: C. 160 mm సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: B. 120 mm సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. 630 mm సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: C. 173.2 MPa సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. The direct load is distributed over the total effective weld length and the eccentric moment produces a secondary shear whose magnitude varies with distance from the weld-group centroid సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. 1.458 kN/mm సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: C. The weld is inadequate because its resistance is only approximately 0.795 kN/mm సమాధానం: C.
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Question 26 ప్రశ్న 26Answer: A. Welded connections generally avoid fastener holes and can provide continuous force transfer, but weld quality, residual stresses, distortion, fatigue behaviour and fabrication control still require careful consideration సమాధానం: A.
