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This section includes 392 Mcqs, each offering curated multiple-choice questions to sharpen your Civil Engineering knowledge and support exam preparation. Choose a topic below to get started.
301. |
A simply supported beam which carries a uniformly distributed load has two equal overhangs. To have maximum B.M. produced in the beam least possible, the ratio of the length of the overhang to the total length of the beam, is |
A. | 0.207 |
B. | 0.307 |
C. | 0.407 |
D. | 0.508 |
Answer» B. 0.307 | |
302. |
A compound truss may be formed by connecting two simple rigid frames, by |
A. | two bars |
B. | three bars |
C. | three parallel bars |
D. | three bars intersecting at a point. |
Answer» C. three parallel bars | |
303. |
A square column carries a load P at the centroid of one of the quarters of the square. If a is the side of the main square, the combined bending stress will be |
A. | 300 mm |
B. | 350 mm |
C. | 400 mm |
Answer» D. | |
304. |
A steel rod 1 metre long having square cross section is pulled under a tensile load of 8 tonnes. The extension in the rod was 1 mm only. If Esteel = 2 x 106 kg/cm2, the side of the rod, is |
A. | 18 m |
B. | 1 cm |
C. | 1.5 cm |
D. | 2 cm |
Answer» D. 2 cm | |
305. |
In plastic analysis, the shape factor for a triangular section, is |
A. | |
B. | 1.5 |
C. | 1.34 |
Answer» D. | |
306. |
The ratio of maximum and average shear stresses on a rectangular section, is |
A. | |
B. | 1 |
C. | 1.25 |
Answer» D. | |
307. |
A concentrated load P is supported by the free end of a quadrantal ring AB whose end B is fixed. The ratio of the vertical to horizontal deflections of the end A, is |
A. | indeterminate |
B. | π |
Answer» C. | |
308. |
The force in AD of the truss shown in given figure, is |
A. | sin2θ |
B. | 4.0t compression |
C. | 3.0t compression |
D. | 0.5t compression |
Answer» D. 0.5t compression | |
309. |
The ratio of crippling loads of a column having both the ends fixed to the column having both the ends hinged, is |
A. | 1 |
B. | 2 |
C. | 3 |
D. | 4 |
Answer» E. | |
310. |
A shaft rotating N.R.M. under a torque T, transmits a power |
A. | between B and C |
B. | Newton metres/sec |
C. | Newton metres/min |
D. | Newton metres/min |
Answer» B. Newton metres/sec | |
311. |
The ratio of tangential and normal components of a stress on an inclined plane through θ° to the direction of the force, is : |
A. | sin θ |
B. | cos θ |
C. | tan θ |
D. | cos θ |
Answer» D. cos Œ∏ | |
312. |
A three hinged arch is generally hinged at its supports and |
A. | at one quarter span |
B. | at the crown |
C. | any where in the rib |
D. | none of these. |
Answer» D. none of these. | |
313. |
For beams breadth is constant, |
A. | depth d ‚àù M |
B. | depth d ‚àù M |
C. | depth d ‚àù 3 M |
D. | depth d ‚àù |
Answer» C. depth d ‚àù 3 M | |
314. |
The locus of the end point of the resultant of the normal and tangential components of the stress on an inclined plane, is |
A. | circle |
B. | parabola |
C. | ellipse |
D. | straight line. |
Answer» D. straight line. | |
315. |
The yield moment of a cross section is defined as the moment that will just produce the yield stress in |
A. | the outer most fibre of the section |
B. | the inner most fibre of the section |
C. | the neutral fibre of the section |
D. | the fibre everywhere |
Answer» B. the inner most fibre of the section | |
316. |
The horizontal thrust on the ends of a two hinged semicircular arch of radius R carrying |
A. | a uniformly distributed load ω per unit run over its right half span, is |
B. | a uniformly distributed load ω per unit run over its entire span is |
C. | a distributed load varying from zero at the left end to ω per unit horizontal run at the right end, is |
D. | all the above. |
Answer» E. | |
317. |
If a three hinged parabolic arch, (span l, rise h) is carrying a uniformly distributed load w/unit length over the entire span, |
A. | horizontal thrust is wl2/8h |
B. | S.F. will be zero throughout |
C. | B.M. will be zero throughout |
D. | all the above. |
Answer» E. | |
318. |
For determining the force in the member AB of the truss shown in the given figure by method of sections, the section is made to pass through AB, AD and ED and the moments are taken about |
A. | joint C |
B. | joint B |
C. | joint D |
D. | joint A |
Answer» D. joint A | |
319. |
For calculating the permissible stress is the emprical formula, known as |
A. | Straight line formula |
B. | Parabolic formula |
C. | Perry's formula |
D. | Rankine's formula. |
Answer» E. | |
320. |
A steel bar 5 m x 50 mm is loaded with 250, 000 N. If the modulus of elasticity of the material is 0.2 MN/mm2 and Poisson's ratio is 0.25, the change in the volume of the bar is : |
A. | 1.125 cm3 |
B. | 2.125 cm3 |
C. | 3.125 cm3 |
D. | 4.125 cm2 |
Answer» D. 4.125 cm2 | |
321. |
A steel rod of sectional area 250 sq. mm connects two parallel walls 5 m apart. The nuts at the ends were tightened when the rod was heated to 100°C. If αsteel = 0.000012/C°, Esteel = 0.2 MN/mm2, the tensile force developed at a temperature of 50°C, is |
A. | 80 N/mm2 |
B. | 100 N/mm2 |
C. | 120 N/mm2 |
D. | 150 N/mm2 |
Answer» D. 150 N/mm2 | |
322. |
The maximum height of a masonry dam of a triangular section whose base width is b and specific gravity s, is |
A. | bs |
B. | b.s |
C. | bs |
D. | s b |
Answer» B. b.s | |
323. |
Slenderness ratio of a long column, is |
A. | area of cross-section divided by radius of gyration |
B. | area of cross-section divided by least radius of gyration |
C. | radius of gyration divided by area of cross-section |
D. | length of column divided by least radius of gyration. |
Answer» E. | |
324. |
The equivalent length of a column of length L, having both the ends hinged, is |
A. | 2L |
B. | L |
Answer» C. | |
325. |
In the truss shown in given figure, the force in member BD is |
A. | 100 t compressive |
B. | 100 t tensile |
C. | zero |
D. | indeterminate |
Answer» B. 100 t tensile | |
326. |
A simply supported rolled steel joist 8 m long carries a uniformly distributed load over it span so that the maximum bending stress is 75 N/mm2. If the slope at the ends is 0.005 radian and the value of E = 0.2 x 106 N/mm2, the depth of the joist, is |
A. | 200 mm |
B. | 250 mm |
C. | 300 mm |
D. | 350 mm |
Answer» E. | |
327. |
For calculating the allowable stress of long columns is the empirical formula, known as |
A. | Straight line formula |
B. | Parabolic formula |
C. | Perry's formula |
D. | Rankine's formula. |
Answer» C. Perry's formula | |
328. |
The force in AC of the truss shown in the given figure, is |
A. | 5t tension |
B. | 4t tension |
C. | 4t compression |
D. | 5t compression |
Answer» C. 4t compression | |
329. |
Maximum strain theory for the failure of a material at the elastic limit, is known as |
A. | Guest's or Trecas' theory |
B. | St. Venant's theory |
C. | Rankine's theory |
D. | Haig's theory |
Answer» C. Rankine's theory | |
330. |
is the equation of Euler's crippling load, if |
A. | both the ends are fixed |
B. | both the ends are hinged |
C. | one end is fixed and other end is free |
D. | one end is fixed and other end is hinged. |
Answer» D. one end is fixed and other end is hinged. | |
331. |
The locus of reaction of a two hinged semi-circular arch, is |
A. | straight line |
B. | parabola |
C. | circle |
D. | hyperbola. |
Answer» B. parabola | |
332. |
The greatest load which a spring can carry without getting permanently distorted, is called |
A. | stiffness |
B. | prof resilience |
C. | proof stress |
D. | proof load. |
Answer» E. | |
333. |
Maximum principal stress theory for the failure of a material at elastic point, is known |
A. | Guest's or Trecas' theory |
B. | St. Venant's theory |
C. | Rankine's theory |
D. | Haig's theory |
Answer» D. Haig's theory | |
334. |
In a shaft, the shear stress is not directly proportional to |
A. | radius of the shaft |
B. | angle of twist |
C. | length of the shaft |
D. | modulus of rigidity. |
Answer» D. modulus of rigidity. | |
335. |
To determine the force in BD of the truss shown in the given figure a section is passed through BD, CD and CE, and the moments are taken about |
A. | A joint |
B. | B joint |
C. | C joint |
D. | D joint. |
Answer» D. D joint. | |
336. |
The locus of the moment of inertia about inclined axes to the principal axis, is |
A. | straight line |
B. | parabola |
C. | circle |
D. | ellipse. |
Answer» E. | |
337. |
The ratio of moments of inertia of a triangular section about its base and about a centroidal axis parallel to its base, is |
A. | 1 |
B. | 1.5 |
C. | 2 |
D. | 2.5 |
Answer» E. | |
338. |
The normal and tangential components of stress on an inclined plane through θ° to the direction of the force, will be equal if θ is |
A. | 45° |
B. | 30° |
C. | 60° |
D. | 90° |
Answer» B. 30¬∞ | |
339. |
At any point of a beam, the section modulus may be obtained by dividing the moment of inertia of the section by |
A. | depth of the section |
B. | depth of the neutral axis |
C. | maximum tensile stress at the section |
D. | maximum compressive stress at the section |
Answer» C. maximum tensile stress at the section | |
340. |
A masonry dam (density = 20, 000 N/m3) 6 m high, one metre wide at the top and 4 m wide at the base, has vertical water face. The minimum stress at the base of the dam when the reservoir is full, will be |
A. | 75 N/m2 |
B. | 750 N/m2 |
C. | 7500 N/m2 |
D. | 75000 N/m2 |
Answer» D. 75000 N/m2 | |
341. |
A close coil helical spring when subjected to a moment M having its axis along the axis of the helix |
A. | it is subjected to pure bending |
B. | its mean diameter will decrease |
C. | its number of coils will increase |
D. | all the above. |
Answer» B. its mean diameter will decrease | |
342. |
The equivalent length of a column of length L having one end fixed and the other end free, is |
A. | 2L |
B. | L |
Answer» B. L | |
343. |
is the equation of Euler's crippling load if |
A. | both at the ends are fixed |
B. | both the ends are hinged |
C. | one end is fixed and other end is free |
D. | one end is fixed and other end is hinged. |
Answer» B. both the ends are hinged | |
344. |
A truss containing j joints and m members, will be a simple truss if |
A. | m = 2j - 3 |
B. | j = 2m - 3 |
C. | m = 3j - 2 |
D. | j = 3m - 2 |
Answer» B. j = 2m - 3 | |
345. |
The tangential component of stress on an plane inclined θ° to the direction of the force, may be obtained by multiplying the normal stress by |
A. | sin θ |
B. | cos θ |
C. | tan θ |
D. | sin θ cos θ |
Answer» E. | |
346. |
For a strongest rectangular beam cut from a circular log, the ratio of the width and depth, is |
A. | 0.303 |
B. | 0.404 |
C. | 0.505 |
D. | 0.606 |
Answer» E. | |
347. |
If a solid shaft (diameter 20 cm, length 400 cm, N = 0.8 x 105 N/mm2) when subjected to a twisting moment, produces maximum shear stress of 50 N/mm2, the angle of twist in radians, is |
A. | 0.001 |
B. | 0.002 |
C. | 0.0025 |
D. | 0.003 |
Answer» D. 0.003 | |
348. |
If normal stresses due to longitudinal and transverse loads on a bar are σ1 and σ2 respectively, the normal component of the stress on an inclined plane θ° to the longitudinal load, is |
A. | σ1 sin θ x σ2 cos θ |
B. | σ1 sin θ2 + σ2 cos2 θ |
Answer» C. | |
349. |
If ΣH and ΣV are the algebraic sums of the forces resolved horizontally and vertically respectively, and ΣM is the algebraic sum of the moments of forces about any point, for the equilibrium of the body acted upon |
A. | ΣH = 0 |
B. | ΣV = 0 |
C. | ΣM = 0 |
D. | all the above. |
Answer» E. | |
350. |
There are two hinged semicircular arches A, B and C of radii 5 m, 7.5 m and 10 m respectively and each carries a concentrated load W at their crowns. The horizontal thrust at their supports will be in the ratio of |
A. | 1 : 1 1/2 : 2 |
B. | 2 : 1 1/2 : 1 |
C. | 0.042384259259259 |
D. | none of these. |
Answer» D. none of these. | |