Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
A uniformly distributed moving load of intensity w covers a specified portion of a beam. If y(x) is the influence-line ordinate for a required response, how is the response obtained?
Explanation:
An elemental load over a length dx is w dx. Its contribution to the selected response is w y(x) dx. Integrating over the loaded region gives the total response as the integral of w y(x) dx. For constant load intensity w, this becomes w multiplied by the algebraic area under the relevant portion of the influence line. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
If a distributed moving load has variable intensity w(x) rather than a constant intensity, which expression gives its effect on a structural response having influence-line ordinate y(x)?
Explanation:
A small load element is dP = w(x)dx. That load element produces a response equal to dP multiplied by the local influence-line ordinate y(x). The elemental response is therefore w(x)y(x)dx. Integration over the loaded portion gives the complete response. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
A simply supported beam has span 12 m and carries a UDL of 8 kN/m over the entire span. Using the influence line for the reaction at the left support A, what is RA?
Explanation:
The influence line for RA is a triangle having base 12 m and height 1 at A. Its area is one-half times 12 times 1, which is 6 m. A full-span UDL response equals the load intensity multiplied by this influence-line area. Thus RA = 8 × 6 = 48 kN. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A simply supported beam has span 12 m. A UDL of 6 kN/m acts only from support A to a point 4 m from A. What is the reaction at A?
Explanation:
For RA, the influence-line ordinate is 1 - x/12. The required loaded influence-line area from x = 0 to x = 4 m is the area of a trapezoid. Its parallel ordinates are 1 and 2/3, so its area is 4(1 + 2/3)/2 = 10/3 m. Multiplying by 6 kN/m gives RA = 20 kN. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
For the beam and partial UDL described in the previous question, what is the reaction at support B?
Explanation:
The total applied load is 6 × 4 = 24 kN. From the influence-line calculation, RA is 20 kN. Vertical equilibrium requires RA + RB = 24 kN. Therefore RB = 4 kN. This also equals the UDL intensity multiplied by the corresponding area under the RB influence line over the first 4 m, so Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
A simply supported beam has span 10 m, and section C is 4 m from support A. A UDL of 8 kN/m covers the entire span. What is the bending moment at C?
Explanation:
For section C, a = 4 m and b = 6 m. The peak ordinate of the bending-moment influence line is ab/L = 2.4 m. The complete moment influence line is triangular, so its area is one-half times 10 times 2.4 = 12 m². Multiplying by 8 kN/m gives MC = 96 kN·m. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
A simply supported beam has span 10 m and section C at 4 m from A. A UDL of 10 kN/m acts only from x = 2 m to x = 7 m, where x is measured from A. What bending moment acts at C?
Explanation:
From x = 2 m to C at x = 4 m, the moment influence-line ordinate rises linearly according to 0.6x. The area of this portion is 3.6 m². From x = 4 m to x = 7 m, the ordinate decreases according to 4 - 0.4x, giving an area of 5.4 m². The total loaded influence-line area is therefore 9.0 m², and MC = 10 × 9 = 90 kN·m. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
A simply supported beam has span 10 m and section C at 4 m from A. Define positive shear as VC = RA minus the downward load lying to the left of C. A UDL of 8 kN/m covers the entire span. What is VC?
Explanation:
The total load is 80 kN, giving RA = 40 kN by symmetry. The load lying to the left of C is 8 × 4 = 32 kN. Hence VC = 40 - 32 = +8 kN. The same result follows from multiplying the UDL intensity by the algebraic area under the shear influence line over the full span. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
For the same 10 m beam with section C at 4 m, a movable UDL of 10 kN/m can cover any favorable portion of the span. What is the maximum positive shear at C?
Explanation:
To maximize positive shear, the UDL should cover only the positive portion of the shear influence line. This positive region extends from C to B. It is triangular with base 6 m and maximum ordinate 0.6 immediately to the right of C. Its area is one-half × 6 × 0.6 = 1.8 m. Multiplying by 10 kN/m gives a maximum positive shear of 18 kN, so Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
For the same beam and section, what is the maximum negative shear at C produced by a movable UDL of 10 kN/m?
Explanation:
For maximum negative shear, the UDL should occupy only the negative portion of the shear influence line. This portion extends from A to C. It forms a triangle with base 4 m and ordinate magnitude 0.4 immediately to the left of C. The algebraic area is -one-half × 4 × 0.4 = -0.8 m. Multiplying by 10 kN/m gives -8 kN, so Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
A simply supported beam has span 10 m and section C at 4 m from A. A movable UDL of 10 kN/m has a fixed loaded length of 4 m. For maximum bending moment at C, how should the UDL be positioned, and what is the resulting MC?
Explanation:
For a fixed-length UDL on the triangular moment influence line, the two end ordinates should be equal at the critical position. This makes the lengths on the two sides of C proportional to a and b. Since a:b = 4:6, the 4 m loaded length divides into 1.6 m left and 2.4 m right. The influence-line area beneath this interval is 7.68 m². Therefore MC = 10 × 7.68 = 76.8 kN·m and Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A simply supported beam has span 10 m and section C at 4 m from A. A movable UDL of 8 kN/m has a fixed loaded length of 6 m. What is the maximum bending moment at C?
Explanation:
The critical UDL position divides the 6 m loaded length in the same ratio as a:b = 4:6. Thus 2.4 m of the UDL lies to the left of C and 3.6 m lies to the right. Integration of the moment influence line over this interval gives an area of 10.08 m². Multiplication by 8 kN/m gives MC = 80.64 kN·m. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A simply supported beam has span 10 m and section C at 4 m from A. A full-span UDL of 6 kN/m acts together with a 30 kN concentrated load located exactly at C. What is the bending moment at C?
Explanation:
The full-span UDL contribution is w multiplied by the complete moment influence-line area. That area is 12 m², so the UDL contributes 6 × 12 = 72 kN·m. The concentrated load at C occupies the peak ordinate of 2.4 m and contributes 30 × 2.4 = 72 kN·m. By superposition, the total moment is 144 kN·m. Therefore Option D is correct.
వివరణ:
Question 14
ప్రశ్న 14
A simply supported beam has span 10 m and section C at x = 4 m. A UDL of 5 kN/m acts from x = 2 m to x = 6 m, and a 20 kN concentrated load acts at x = 8 m. What is MC?
Explanation:
The area under the moment influence line from x = 2 m to x = 6 m is 7.6 m². The UDL therefore contributes 5 × 7.6 = 38 kN·m. At x = 8 m, the moment influence ordinate is 4(10-8)/10 = 0.8 m. The point load contributes 20 × 0.8 = 16 kN·m, giving a total MC of 54 kN·m. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A simply supported beam has span 12 m. A full-span UDL of 4 kN/m acts together with a 30 kN concentrated load located 3 m from support A. What is the reaction at A?
Explanation:
The full-span UDL has total load 48 kN and contributes 24 kN to RA by symmetry. The reaction influence-line ordinate at x = 3 m is 9/12 = 0.75. The 30 kN point load therefore contributes 22.5 kN to RA. The combined reaction is 24 + 22.5 = 46.5 kN. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
A simply supported beam has span 12 m. A 30 kN moving point load and a movable UDL of 5 kN/m having fixed length 4 m may be positioned independently to maximize reaction RA. What is the maximum RA?
Explanation:
The point load maximizes RA when it is located at A, contributing the full 30 kN. The 4 m UDL should also begin at A because the RA influence line decreases toward B. The area under the RA influence line over the first 4 m is 10/3 m. The UDL contribution is therefore 5 × 10/3 = 16.67 kN. The total maximum reaction is approximately 46.67 kN, so Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A simply supported beam has span 10 m and section C at 4 m from A. A 30 kN point load and a 4 m long UDL of 5 kN/m may be positioned independently to maximize bending moment at C. What is the maximum combined MC?
Explanation:
The 30 kN concentrated load should be placed at C, where the moment influence ordinate is 2.4 m. It therefore contributes 72 kN·m. The 4 m UDL is positioned critically with 1.6 m to the left and 2.4 m to the right of C, giving influence-line area 7.68 m². Its contribution is 5 × 7.68 = 38.4 kN·m. Thus the maximum combined moment is 110.4 kN·m and Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
An overhanging beam has supports A at x = 0 and B at x = 6 m, with a right overhang extending to x = 8 m. A UDL of 6 kN/m acts only over the overhang from x = 6 m to x = 8 m. What are the support reactions?
Explanation:
The UDL over the 2 m overhang has a resultant of 12 kN acting at x = 7 m. Taking moments about A gives RB × 6 = 12 × 7. Thus RB = 14 kN upward. Vertical equilibrium then gives RA + 14 - 12 = 0, so RA = -2 kN. The negative sign means RA acts 2 kN downward, making Option B correct.
వివరణ:
Question 19
ప్రశ్న 19
A simply supported beam has span 12 m. A movable UDL of 6 kN/m has fixed length 5 m. What is the maximum reaction at support A?
Explanation:
The RA influence line decreases continuously from 1 at A to zero at B. Therefore a fixed-length UDL maximizes RA when it begins at A. The area beneath the first 5 m of the influence line is 5 - 25/24 = 95/24 m. Multiplying by 6 kN/m gives 23.75 kN. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
A 4 m long UDL of 5 kN/m moves together with a 30 kN concentrated load located at the right-hand end of the UDL. This combined system moves across a simply supported beam of span 12 m. What is the maximum reaction at support B?
Explanation:
The influence line for RB increases toward support B. The combined system therefore produces its maximum RB when the 30 kN load is at B and the UDL occupies the final 4 m of the span. The area under the RB influence line from x = 8 m to x = 12 m is 10/3 m. The UDL contributes 5 × 10/3 = 16.67 kN, while the point load contributes 30 kN. Hence the maximum RB is approximately 46.67 kN and Option D is correct.
వివరణ:
Question 21
ప్రశ్న 21
A simply supported beam has span 10 m and section C at 4 m from A. A 20 kN point load is placed immediately to the right of C, and a UDL of 4 kN/m extends for 3 m from C toward B. Using VC = RA minus loads strictly to the left of C, what is VC?
Explanation:
The shear influence ordinate immediately to the right of C is 0.6, so the 20 kN point load contributes 12 kN. Over the next 3 m, the positive shear influence ordinate decreases linearly from 0.6 to 0.3. The area under that portion is one-half times 3 times the sum 0.6 + 0.3, which is 1.35 m. The UDL contributes 4 × 1.35 = 5.4 kN, giving VC = 17.4 kN. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
For a simply supported beam, the influence line for bending moment at an interior section is positive throughout the span. If a movable UDL has no restriction on loaded length other than remaining on the beam, where should it be placed to maximize the positive bending moment at that section?
Explanation:
A positive UDL contribution is obtained wherever the bending-moment influence-line ordinate is positive. For a simply supported beam and an interior section, the moment influence line is positive throughout the span. Loading a greater positive area therefore increases the response. With no restriction on the loaded length, the maximum positive moment occurs when the UDL covers the entire span. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Consider the following statements about moving UDLs and influence lines:
1. The response to a UDL equals its intensity multiplied by the algebraic influence-line area under the loaded region.
2. To maximize positive shear at a section using an unrestricted movable UDL, the load should occupy the positive region of the shear influence line.
3. A fixed-length UDL producing maximum moment at an interior section must always be centred geometrically about that section.
4. Negative influence-line areas must retain their signs when distributed-load responses are calculated.
Which statements are correct?
1. The response to a UDL equals its intensity multiplied by the algebraic influence-line area under the loaded region.
2. To maximize positive shear at a section using an unrestricted movable UDL, the load should occupy the positive region of the shear influence line.
3. A fixed-length UDL producing maximum moment at an interior section must always be centred geometrically about that section.
4. Negative influence-line areas must retain their signs when distributed-load responses are calculated.
Which statements are correct?
Explanation:
Statement 1 is the standard influence-line area rule for a uniform distributed load. Statement 2 is correct because loading only positive ordinates maximizes a positive response when loaded length is unrestricted. Statement 3 is incorrect because the optimum position of a fixed-length UDL depends on the slopes of the influence line and is generally not geometrically centred about an unsymmetrically located section. Statement 4 is correct because influence-line ordinates represent algebraic response. Therefore Statements 1, 2 and 4 are correct and Option B is the unique answer.
వివరణ:
Question 24
ప్రశ్న 24
Match each influence-line application with its correct principle:
1. Full or partial UDL response
2. Maximum positive section shear under unrestricted movable UDL
3. Fixed-length UDL for maximum moment at an interior section
4. Several combined moving load components
P. Position the fixed loaded length so that its end ordinates on the moment influence line are equal
Q. Add the separate responses by linear superposition
R. Load the positive portion of the shear influence line
S. Multiply load intensity by the relevant algebraic influence-line area
Which matching is correct?
1. Full or partial UDL response
2. Maximum positive section shear under unrestricted movable UDL
3. Fixed-length UDL for maximum moment at an interior section
4. Several combined moving load components
P. Position the fixed loaded length so that its end ordinates on the moment influence line are equal
Q. Add the separate responses by linear superposition
R. Load the positive portion of the shear influence line
S. Multiply load intensity by the relevant algebraic influence-line area
Which matching is correct?
Explanation:
A UDL response is obtained from its intensity multiplied by the relevant influence-line area, so item 1 matches S. Maximum positive shear is obtained by loading the positive influence-line region, so item 2 matches R. For a fixed-length UDL on a piecewise linear moment influence line, the critical position has equal ordinates at its two loaded ends, so item 3 matches P. Combined load effects are added using linear superposition, so item 4 matches Q. Therefore Option A gives the correct complete matching.
వివరణ:
Question 25
ప్రశ్న 25
A simply supported beam has span 10 m and section C at 4 m from A. Define VC = RA minus the downward loads strictly to the left of C. A UDL of 6 kN/m acts from x = 0 to x = 2 m, while another UDL of 4 kN/m acts from x = 6 m to x = 10 m. What is VC?
Explanation:
Over x = 0 to 2 m, the shear influence line lies on the negative branch and its algebraic area is -0.2 m. The first UDL therefore contributes 6(-0.2) = -1.2 kN. From x = 6 to 10 m, the positive influence-line area is 0.8 m. The second UDL contributes 4(0.8) = 3.2 kN. The total shear is -1.2 + 3.2 = +2.0 kN, so Option C is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: A. By multiplying w by the algebraic area under the influence line over the loaded length సమాధానం: A.
-
Question 2 ప్రశ్న 2Answer: A. Response = ∫w(x)y(x) dx over the loaded region సమాధానం: A.
-
Question 3 ప్రశ్న 3Answer: B. 48 kN సమాధానం: B.
-
Question 4 ప్రశ్న 4Answer: C. 20 kN సమాధానం: C.
-
Question 5 ప్రశ్న 5Answer: A. 4 kN సమాధానం: A.
-
Question 6 ప్రశ్న 6Answer: C. 96 kN·m సమాధానం: C.
-
Question 7 ప్రశ్న 7Answer: C. 90 kN·m సమాధానం: C.
-
Question 8 ప్రశ్న 8Answer: C. +8 kN సమాధానం: C.
-
Question 9 ప్రశ్న 9Answer: C. 18 kN సమాధానం: C.
-
Question 10 ప్రశ్న 10Answer: B. -8 kN సమాధానం: B.
-
Question 11 ప్రశ్న 11Answer: B. 1.6 m to the left and 2.4 m to the right of C; 76.8 kN·m సమాధానం: B.
-
Question 12 ప్రశ్న 12Answer: C. 80.64 kN·m సమాధానం: C.
-
Question 13 ప్రశ్న 13Answer: D. 144 kN·m సమాధానం: D.
-
Question 14 ప్రశ్న 14Answer: C. 54 kN·m సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: C. 46.5 kN సమాధానం: C.
-
Question 16 ప్రశ్న 16Answer: C. 46.67 kN సమాధానం: C.
-
Question 17 ప్రశ్న 17Answer: C. 110.4 kN·m సమాధానం: C.
-
Question 18 ప్రశ్న 18Answer: B. RA = 2 kN downward and RB = 14 kN upward సమాధానం: B.
-
Question 19 ప్రశ్న 19Answer: C. 23.75 kN సమాధానం: C.
-
Question 20 ప్రశ్న 20Answer: D. 46.67 kN సమాధానం: D.
-
Question 21 ప్రశ్న 21Answer: C. 17.4 kN సమాధానం: C.
-
Question 22 ప్రశ్న 22Answer: A. Over the entire span సమాధానం: A.
-
Question 23 ప్రశ్న 23Answer: B. 1, 2 and 4 only సమాధానం: B.
-
Question 24 ప్రశ్న 24Answer: A. 1-S, 2-R, 3-P, 4-Q సమాధానం: A.
-
Question 25 ప్రశ్న 25Answer: C. +2 kN సమాధానం: C.
