Civilwar Day 5(Sep 19)-PWD

A ratio of moment carrying capacity of a circular beam of diameter D and square beam of size D is:

• π/4
• 3π/8
• π/3
• 3π/16

Answer explanation

1. Circular beam (diameter D):
   I = (π D^4) / 64
   ymax = D/2
   Zc = (π D^4 / 64) / (D/2) = (π D^3) / 32

2. Square beam (side D):
   I = D^4 / 12
   ymax = D/2
   Zs = (D^4 / 12) / (D/2) = D^3 / 6

3. Ratio:
   Zc / Zs = (π D^3 / 32) / (D^3 / 6) = 3π / 16

Final Answer: 3π / 16

Pure bending is that bending where

• Shear stress is maximum
• Bending stress is maximum
• Shear stress is nil
• Bending stress is nil

Answer explanation

Pure bending occurs with constant moment and no shear force, hence shear stress is nil.

Which of the following is the most efficient in carrying the bending moment?

• Rectangular section
• Circular section
• I-Section
• T-section

Answer explanation

I-section is most efficient since material is placed farthest from the neutral axis.

Beam of uniform strength means

• beam in which tensile stress developed is constant
• beam in which bending stress developed is constant
• is equal to the allowable stress
• Both b and c

Answer explanation

Beam of uniform strength has constant bending stress, equal to allowable stress.

In a simply supported beam loaded with U.D.L over the whole section, the bending stress is …………. at top and ………….. at bottom.

• Compressive, tensile
• Tensile, compressive
• Tensile, zero
• Compressive, zero

Answer explanation

Top fibers are in compression and bottom fibers are in tension under UDL.

Section modulus of hollow rectangular section is

• (BD³ − bd³)/6D
• (BD³ − bd³)/6B
• (BD³ − bd³)/12D
• (BD³ − bd³)/12B

Answer explanation

Z = I/ymax, with I = (BD³−bd³)/12 and ymax = D/2 gives Z = (BD³−bd³)/6D.

In an I -section , the bending moment is resisted mainly by

• Flanges only
• Web only
• Both by flanges and web
• None of these

Answer explanation

Flanges resist bending moment; web carries shear force.

A rectangular beam of size 60 mm x 100 mm has a central rectangular hole of size 15 mm x 20 mm. The beam is subjected to bending and the maximum bending stress is limited to 100 N/mm² . Find the moment of resistance of the hollow beam section.

• 30kNm

• 20kNm

• 10kNm

• 5kNm

Answer explanation

Given:

Outer rectangle = 60 mm × 100 mm

Central hole = 15 mm × 20 mm

Allowable bending stress = 100 N/mm²

Step 1: Moment of inertia

I = Iouter – Ihole

Iouter = (b h³) / 12 = (60 × 100³) / 12 = 5,000,000 mm⁴

Ihole = (15 × 20³) / 12 = (15 × 8000) / 12 = 10,000 mm⁴

I = 5,000,000 – 10,000 = 4,990,000 mm⁴

Step 2: Section modulus

ymax = 100 / 2 = 50 mm

Z = I / ymax = 4,990,000 / 50 = 99,800 mm³

Step 3: Moment of resistance

M = σ × Z = 100 × 99,800 = 9.98 × 10⁶ N·mm

= 9.98 × 10³ N·m

Final Answer:

Moment of resistance = 9.98 × 10⁶ N·mm (≈ 9.98 kN·m)

A beam strongest in flexure is the one which has

• maximum bending stress
• maximum area of cross section
• maximum section modulus
• maximum moment of inertia

Answer explanation

Strength in flexure depends on section modulus.

A simply supported beam of span 6m and 75 mm diameter carries a udl of 1.5 KN/m. What is the maximum value of bending stress (Mpa) ?

• 162.98 MPa
• 325.95 MPa
• 625.95 MPa
• 651.90 MPa

Answer explanation

Given:

Span L = 6 m = 6000 mm

Diameter d = 75 mm

UDL w = 1.5 kN/m = 1500 N/m = 1.5 N/mm

Step 1: Maximum bending moment

M = wL² / 8

= (1.5 × 6000²) / 8

= (1.5 × 36,000,000) / 8

= 6.75 × 10^6 N·mm

Step 2: Section modulus for circular section

Z = (π d³) / 32

= (3.1416 × 75³) / 32

= 41,428 mm³

Step 3: Bending stress

σ = M / Z

= (6.75 × 10^6) / 41,428

≈ 163 N/mm²

Answer: Maximum bending stress = 163 MPa