Past Year Questions: Bending & Shear Stresses

# Past Year Questions: Bending & Shear Stresses | Solid Mechanics - Mechanical Engineering PDF Download

Q.1  For a channel section subjected to a downward vertical shear force at its centroid, which one of the following represents the correct distribution of shear stress in flange and web?    [2019 : 1 Mark, Set-ll]
(a)

(b)

(c)

(d)

Ans.
(C)
Solution:
Shear flow distribution for channel.

Q.2 Cross section of a built-up wooden beam as shown in figure (not drawn to scale) is subjected to a vertical shear force of 8 kN. The beam is symmetrical about the neutral axis (NA), shown, and the moment of inertia about N.A. is 1.5 x 109 mm4. Considering that the nails at the location P are spaced longitudinally (along the length of the beam) at 60 mm, each of the nails at P will be subjected to the shear force of    [2019 : 2 Marks, Set-I]

(a) 240 N
(b) 480 N
(c) 60 N
(d) 120 N
Ans.
(A)
Solution:

Shear Flow,

Distance between two nails l = 60 mm
∴ S.F. resisted by each nail = q x l = 240 N

Q.3 For a given loading on a rectangular plain concrete beam with an overall depth of 500 mm, the compressive strain and tensile strain developed at the extreme fibers are of the same magnitude of 2.5 x 10-4. The curvature in the beam cross-section (in m-1, round off to 3 decimal places), is _______.    [2019 : 1 Mark, Set-I]
Solution:

Given: D= 500 mm

= 1*10-6

Q.4 An 8 m long simply-supported elastic beam of rectangular cross-section (100 mm x 200 mm) is subjected to a uniformly distributed load of 10kN/m over its entire span. The maximum principal stress (in MPa, up to two decimal places) at a point located at the extreme compression edge of a cross-section and at 2 m from the support is ______ .    [2018 : 2 Marks, Set-II]
Solution:

[Due to symmetry]
MA = (-10 x 2 x 1) + 40 x 2
= 60 kNm

σ = My/I

Direct shear stress = 0
Principal stress,

So principal stress
= 90 N/mm2 = 90 MPa

Q.5 A cantilever beam of length 2 m with a square section of side length 0.1 m is loaded vertically at the free end. The vertical displacement at the free end is 5 mm. The beam is made of steel with Young’s modulus of 2.0 x 1011 N/m2. The maximum bending stress at the fixed end of the cantilever is    [2018 : 2 Marks, Set-I]
(a) 20.0 MPa
(b) 37.5 MPa
(c) 60.0 MPa
(d) 75.0 MPa
Ans.
(B)
Solution:

= 37.5 x 106 N/m2 = 37.5 MPa

Q.6 A 450 mm long plain concrete prism is subjected to the concentrated vertical loads as shown in the figure. Cross section of the prism is given as 150 mm x 150 mm. Considering linear stress distribution across the cross-section, the modulus of rupture (expressed in MPa) is ___ .    [2016 : 2 Marks, Set-II]

Solution:

BMQ = 11.25 x 150
= 1.6875 x 106 N-mm

where,

Q.7 A simply supported reinforced concrete beam of length 10 m sags while undergoing shrinkage. Assuming a uniform curvature of 0.004 m-1 along the span, the maximum deflection (in m) of the beam at mid-span is ______ .    [2015 : 2 Marks, Set-II]
Solution: Method - I

= 249.95 m
Deflection = AA' = 250 - 249.95
= 0.05 m
Method-ll

δ 0.05 m
= 71.12N/mm2

Q.8 A symmetric l-section (with width of each flange = 50 mm, thickness of web = 10 mm) of steel is subjected to a shear force of 100 kN. Find the magnitude of the shear stress (in N/mm2) in the web at its junction with the top flange _____ .    [2013 : 1 Mark]
Solution:

Q.9 The “Plane section remain plane” assumption in bending theory implies    [2013 : 1 Mark]
(a) strain profile is linear
(b) stress profile is linear
(c) both profiles are linear
(d) shear deformation is neglected
Ans.
(A)
Solution:

Since, E ∝ δy
So, strain varies linearly.

Q.10 Consider a simply supported beam with a uniformly distributed load having a neutral axis (NA) as shown. For points P(on the neutral axis) and Q (at the bottom of the beam) the state of stress is best represented by which of the following pairs?    [2011 : 1 Mark]

Ans.
(B)
Solution:

Point P: Point Plies on NA, hence bending stress is zero at point P.
Point Palso lies at mid span, so shear force, V = 0
⇒ Shear stress, τ = 0
∴ State of stress of point Pwill be,

Point Q: At point Q flexural stress is maximum and nature of which is tensile due to downward loading. Point Q lies at the extreme of beam, therefore, shear stress at point Q is zero.
∴ State of stress of point Q will be,

The document Past Year Questions: Bending & Shear Stresses | Solid Mechanics - Mechanical Engineering is a part of the Mechanical Engineering Course Solid Mechanics.
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## Solid Mechanics

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## FAQs on Past Year Questions: Bending & Shear Stresses - Solid Mechanics - Mechanical Engineering

 1. What is bending stress and how is it calculated?
Ans. Bending stress is a type of stress that occurs in a structural member when it is subjected to bending moments. It is calculated using the formula: Bending Stress = (M * y) / I where M is the bending moment applied, y is the distance from the neutral axis to the point under consideration, and I is the moment of inertia of the cross-sectional area.
 2. What is shear stress and how is it calculated?
Ans. Shear stress is a type of stress that occurs when forces are applied parallel or tangential to the surface of a material. It is calculated using the formula: Shear Stress = (F * A) / (t * l) where F is the force applied parallel to the surface, A is the cross-sectional area, t is the thickness of the material, and l is the length of the material.
 3. What is the difference between bending stress and shear stress?
Ans. The main difference between bending stress and shear stress is the direction in which the forces are applied. Bending stress occurs when forces are applied perpendicular to the longitudinal axis of a member, causing it to bend. Shear stress, on the other hand, occurs when forces are applied parallel or tangential to the surface of a material, causing it to deform or shear.
 4. How do bending and shear stresses affect the strength of a material?
Ans. Bending stress and shear stress both contribute to the overall stress state of a material. They can cause deformation, failure, or structural instability depending on the magnitude and distribution of the stresses. Excessive bending stress can lead to bending failure or fracture, while excessive shear stress can cause shear failure or deformation.
 5. What are some common applications of bending and shear stresses in engineering?
Ans. Bending and shear stresses are encountered in various engineering applications. Some common examples include the design of beams, columns, and bridges, where bending stresses play a critical role in determining structural integrity. Shear stresses are important in the design of mechanical joints, bolts, and fasteners, where they affect the strength and reliability of connections.

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