If ‘p’ is the net upward pressure on a square footing of side ‘b’ for a square column of side ‘a’, the maximum bending moment is given by

If q is the punching shear resistance per unit area a, is the side of a square footing for a column of side b, carrying a weight W including the weight of the footing, the depth (D) of the footing from punching shear consideration, is

A

$${\text{D}} = \frac{{{\text{W}}\left( {{\text{a}} - {\text{b}}} \right)}}{{4{{\text{a}}^2}{\text{bq}}}}$$

B

$${\text{D}} = \frac{{{\text{W}}\left( {{{\text{a}}^2} - {{\text{b}}^2}} \right)}}{{4{{\text{a}}^2}{\text{bq}}}}$$

C

$${\text{D}} = \frac{{{\text{W}}\left( {{{\text{a}}^2} - {{\text{b}}^2}} \right)}}{{8{{\text{a}}^2}{\text{bq}}}}$$

D

$${\text{D}} = \frac{{{\text{W}}\left( {{{\text{a}}^2} - {{\text{b}}^2}} \right)}}{{4{\text{abq}}}}$$

If W is total load per unit area on a panel, D is the diameter of the column head, L is the span in two directions, then the sum of the maximum positive bending moment and average of the negative bending moment for the design of the span of a square flat slab, should not be less than

A

$$\frac{{{\text{WL}}}}{{12}}{\left( {{\text{L}} - \frac{{2{\text{D}}}}{3}} \right)^2}$$

B

$$\frac{{{\text{WL}}}}{{10}}{\left( {{\text{L}} + \frac{{2{\text{D}}}}{3}} \right)^2}$$

C

$$\frac{{{\text{WL}}}}{{10}}{\left( {{\text{L}} - \frac{{2{\text{D}}}}{3}} \right)^2}$$

D

$$\frac{{{\text{WL}}}}{{12}}{\left( {{\text{L}} - \frac{{\text{D}}}{3}} \right)^2}$$

A shaft of diameter D is subjected to a twisting moment (T) and a bending moment (M). If the maximum bending stress is equal to maximum shear stress developed, then M is equal to

A

$$\frac{{\text{T}}}{2}$$

B

T

C

2T

D

4T

The bending moment at a section tends to bend or deflect the beam and the internal stresses resist its bending. The resistance offered by the internal stresses, to the bending, is called

A

Compressive stress

B

Shear stress

C

Bending stress

D

Elastic modulus

In a combined footing for two columns carrying unequal loads, the maximum hogging bending moment occurs at

A

Less loaded column

B

More loaded column

C

A point of the maximum shear force

D

A point of zero shear force

A shaft is subjected to bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in the shaft, is

A

$$\frac{{\text{M}}}{{\text{T}}}$$

B

$$\frac{{\text{T}}}{{\text{M}}}$$

C

$$\frac{{2{\text{M}}}}{{\text{T}}}$$

D

$$\frac{{2{\text{T}}}}{{\text{M}}}$$

The bending moment ‘M’ and a torque ‘T’ is applied on a solid circular shaft. If the maximum bending stress equals to maximum shear stress developed, then ‘M’ is equal to

A

$$\frac{{\text{T}}}{2}$$

B

1T

C

2T

D

4T

When a shaft is subjected to combined twisting moment (T) and bending moment (M), then the equivalent twisting moment is equal to

A

$$\frac{1}{2}\sqrt {{{\text{M}}^2} + {{\text{T}}^2}} $$

B

$$\sqrt {{{\text{M}}^2} + {{\text{T}}^2}} $$

C

$$\frac{1}{2}\left( {{\text{M}} + \sqrt {{{\text{M}}^2} + {{\text{T}}^2}} } \right)$$

D

$${\text{M}} + \sqrt {{{\text{M}}^2} + {{\text{T}}^2}} $$

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

$$\frac{{\text{P}}}{{{{\text{a}}^2}}}$$

B

$$\frac{{2{\text{P}}}}{{{{\text{a}}^2}}}$$

C

$$\frac{{3{\text{P}}}}{{{{\text{a}}^2}}}$$

D

$$\frac{{4{\text{P}}}}{{{{\text{a}}^2}}}$$

In a simple bending theory, one of the assumptions is that the plane sections before bending remain plane after bending. This assumption means that

A

Stress is uniform throughout the beam

B

Strain is uniform throughout the beam

C

Stress is proportional to the distance from the neutral axis

D

Strain is proportional to the distance from the neutral axis

If ‘p’ is the net upward pressure on a square footing of side ‘b’ for a square column of side ‘a’, the maximum bending moment is given by

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