CIVIL WAR DAY 3 (SEP 17)-PWD

If the shear force diagram of a simply supported beam is parabolic, the load on the beam is

• UDL
• Concentrated load
• Uniformly varying load
• Concentrated moment

Answer explanation

A parabolic shear force diagram indicates that the load intensity varies linearly, i.e., a uniformly varying load (UVL).

When there is a sudden increase or decrease in shear force diagram between any two points, it indicates that there is

• No load between the points
• Point load between the points
• UDL between the points
• Concentrated moment between the points

Answer explanation

A sudden jump in the shear force diagram corresponds to a concentrated (point) load acting at that location.

Maximum bending moment of a simply supported beam loaded with UDL w/m throughout and Point load P at mid point is

• PL/4
• wL²/8
• 0
• wL²/8 + PL/4

Answer explanation

For a simply supported beam: BM due to UDL = wL²/8, BM due to point load at midspan = PL/4. Hence, maximum BM = wL²/8 + PL/4.

The rate of change of bending moment is equal to

• Shear force
• Deflection
• Slope
• Axial thrust

Answer explanation

By structural analysis, dM/dx = V, i.e., the rate of change of bending moment with respect to length equals shear force.

A simply supported beam carries two equal loads W at a distance of L/3 from either supports. The maximum bending moment is

• WL/3
• WL/4
• WL/8
• 2WL/3

Answer explanation

Reactions are equal, RA = RB = W.

Maximum bending moment occurs directly under the loads.

M = RA × (L/3) = W × (L/3) = WL/3

The point in a beam where the shear force changes sign is the point of

• Maximum BM
• Contraflexure
• Zero curvature
• Maximum curvature

Answer explanation

Shear force changes sign at the section of maximum bending moment.

A simply supported beam carries a UDL of 10 kN/m for a length of 2 m from the left support. If the total length of the beam is 4 m, the point of maximum bending moment from the left support is at a distance of:

• 1 m
• 1.5 m
• 2 m
• 2.5 m

Answer explanation

Total UDL load = 10 × 2 = 20 kN acting at 1 m from left support.

Reactions: RA = 15 kN, RB = 5 kN.

Bending moment equation within loaded span (0 ≤ x ≤ 2):

M(x) = 15x – 5x².

Differentiating: dM/dx = 15 – 10x = 0 → x = 1.5 m.

Hence, maximum BM occurs at 1.5 m from left support.

How many points of contra-flexure can be there in a simply supported beam

• Three
• None
• Two
• One

Answer explanation

A point of contraflexure is where bending moment changes sign. In a simply supported beam, BM is always positive (or zero at supports). Hence, no contraflexure point exists.

A cantilever beam of length of 2m carries a UDL of 100 N/m over its whole span. The maximum shear force in the beam will

• 200 N
• 400 N
• 100 N
• 200 NM

Answer explanation

For a cantilever with UDL w over span L, max shear at the support = w × L. Here, 100 × 2 = 200 N.

The slope of shear force line at any section of the beam is also called

• Bending moment at that section
• Maximum Shear force
• Rate of loading at that section
• Maximum bending moment

Answer explanation

The slope of the shear force diagram at any section equals the intensity of loading at that section (w = dV/dx).