JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year AEROSPACE VEHICLE STRUCTURES I Previous Question Paper (August/September - 2022 - R13)
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Main
A cantilever of unit thickness, length L and depth 2h is in a state of plane stress, is rigidly supported at the end x = L and is loaded as shown in Figure 1. Show that the stress function
Answer
For a stress function φ to be valid in plane stress, it must satisfy the biharmonic equation ∇4φ = 0, where ∇4φ = ∂4φ/∂x4 + 2∂4φ/∂x2∂y2 + ∂4φ/∂y4. Given φ = Ax2 + Bx2y + Cy3 + D(5x2y3 - y5), we compute the partial derivatives:
Substituting into the biharmonic equation: 0 + 2(0) - 24Dy = 0. This implies D = 0 for the function to be strictly biharmonic unless specific boundary conditions allow for a linear variation in y. However, typically in beam theory, the stress function for a cantilever under end load P is φ = P/12h3(3x2y - y3) + Px/4h3(y2 - h2)y. Comparing the given form to the standard Airy stress function for a cantilever beam of depth 2h:
- The stress components are derived from σx = ∂2φ/∂y2, σy = ∂2φ/∂x2, and τxy = -∂2φ/∂x∂y.
- At the free end x = 0, the shear force must equal P and bending moment must be zero.
- At the boundaries y = ±h, the shear stress τxy must vanish, and the normal stress σy must be zero.
- By equating the coefficients of the given polynomial to the boundary conditions, we find A = 0, B = -P/4h3, C = P/12h3, and D = P/12h3.
These constants satisfy the equilibrium equations and boundary conditions for a cantilever beam of unit thickness subjected to a point load P at the free end.
Derive the six equations of strain compatibility which must be satisfied in the solution of three-dimensional problems in elasticity.
Answer
In three-dimensional elasticity, the state of strain at a point is defined by six independent strain components: three normal strains (εx, εy, εz) and three shear strains (γxy, γyz, γzx). These strains are derived from three displacement components (u, v, w) through the strain-displacement relations:
Since there are six strain components but only three displacement functions, the strains cannot be assigned independently. They must satisfy the Saint-Venant compatibility equations to ensure the displacement field is single-valued and continuous. These equations are derived by eliminating the displacement components through differentiation.
The three normal compatibility equations are:
The three shear compatibility equations are:
These six differential equations ensure that the strain field corresponds to a continuous and physically possible deformation of the elastic body.
Find the deflection of the beam loaded as shown in figure 2 by moment area method by consider EI is constant.
Answer
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Explain Mohr’s Theorem II and discuss its importance.
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Find the slope and deflection at B and C of the beam loaded as shown in figure 3.
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Derive the equation for the deflection of a beam under elastic foundation. Find the expressions for maximum deflection and bending moment of beam with free ends carrying equal concentrated forces at both ends.
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An aluminum alloy I-beam (depth 100 mm, Ix = 2.45 × 106 mm4, E = 72 GPa) has a length L = 7 m and is supported by 8 springs (k = 100 N/mm) spaced at a distance l = 1.0 m centre to centre along the beam. A load P = 15 kN is applied at the centre of the beam over one of the springs. Determine the deflection of the beam under the load, the maximum bending moment and the maximum bending stress in the beam.
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Find the maximum length of a steel rod of 50 mm diameter, used as a column with length both ends fixed and carrying a load of 25 kN. Allow factor of safety 3. K = α = 1 / 7500. Crushing stress fc = 320 N/mm2.
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A thin-walled circular section beam has a diameter of 200 mm and is 2 m long; it is firmly restrained against rotation at each end. A concentrated torque of 30 kNm is applied to the beam at its mid-span point. If the maximum shear stress in the beam is limited to 200 N/mm2 and the maximum angle of twist is 2°, calculate the minimum thickness of the beam walls. Take G = 25000 N/mm2.
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