Test: Bending & Shear Stress - 1 - Civil Engineering (CE) MCQ

Let mid span depth is d. Then BM at a distance L/2 from one end.
∴  
∴ 
∴ 
or 

As we know,

∴ 
Now, 
∴ 
⇒ 
⇒ 

The maximum shear stress in a rectangular beam occurs at NA and is given by,

∴ 
= 1.5 N/mm²

In the flitched beams there are two principles to be followed in analysis,
(i) strain diagram is linear
(ii) The radius of curvature of both materials is same at all the points.

• Allowable stress in weak material = f
• Allowable stress in strong material = 10 f
• Modulus of elasticity of weak material = E
• Modulus of elasticity of strong material = 10E
• Maximum strain in weak material = f/E
• Maximum strain in strong material =10f/10E = f/E
• The hatched material can be replaced by equivalent weak material when its width is multiplied by modular ratio. So the equivalent sections are:

It is very clear that section modulus of section-ll is more than that of section-Ill as more area is close to the neutral axis in section-lI. Section-IV is unequal and its section modulus in tension and compression will be different. Since the strain is (f/E) at topmost and at bottom most fibre it is less than f/E.
So its capacity will be less than that of section-ll. Comparing section-l and section-ll, more area is away from the neutral axis in section-ll, so the moment capacity of section-ll is more than that of section-l. Thus section-ll has higher capacity.

Z = I/(y(max)) = bd2 /6
Za = (200) (600)2 / 6 
Zb = (800) (600)2/6
Za / Zb = 25

• Beam with constant moment of inertia will be prismatic beam.
• Beam in which bending stress is uniform at maximum bending moment cross-section will form a plastic hinge at that section..
• Beam of uniform strength are of two types,
  (i) Beams of constant width
  (ii) Beams of constant depth