The bending moment on a section is maximum where shear force is

A zero
B minimum
C maximum
D changing sing

Correct Answer: zero

The maximum bending moment occurs in a beam when the shear force at that section is zero or changes the sign because at point of contra flexure the bending moment is zero. Explanation: The positive bending moment in a section is considered because it causes convexity downwards.
Bissoy MCQ
#সাধারণ বিজ্ঞান #কাজ #Bangladesh Bank #Assistant Director (Engineering -Mechanical) | 11.01.2019 #ক্ষমতা ও শক্তি

Related Questions

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
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
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
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 $$\frac{{{\text{WL}}}}{{12}}{\left( {{\text{L}} - \frac{{\text{D}}}{3}} \right)^2}$$
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 $${\text{M}} + \sqrt {{{\text{M}}^2} + {{\text{T}}^2}} $$
‘P’ is the pre-stressed force applied to the tendon of a rectangular pre-stressed beam whose area of cross section is ‘A’ and sectional modulus is ‘Z’. The maximum stress ‘f’ in the beam, subjected to a maximum bending moment ‘M’, is
A $${\text{f}} = \frac{{\text{P}}}{{\text{A}}} - \frac{{\text{Z}}}{{\text{M}}}$$
B $${\text{f}} = \frac{{\text{A}}}{{\text{P}}} - \frac{{\text{M}}}{{\text{Z}}}$$
C $${\text{f}} = \frac{{\text{P}}}{{\text{A}}} - \frac{{\text{M}}}{{\text{Z}}}$$
D $${\text{f}} = \frac{{\text{P}}}{{\text{A}}} - \frac{{\text{M}}}{{6{\text{Z}}}}$$
A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be
A $$\frac{\tau }{2}$$
B $$\tau $$
C $$2\tau $$
D $$4\tau $$
A beam of T-section is subjected to a shear force of F. The maximum shear force will occur at the
A Top of the section
B Bottom of the section
C Neutral axis of the section
D Junction of web and flange
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

Next steps