JNTU
JNTU B.Tech III FINITE ELEMENT METHODS Previous Question Paper (November/December - 2020 - R16)
Answer any five questions. All questions carry equal marks.
Main
What category and applications fall under Plane strain and Plane stress problems? Illustrate the same.
Answer
In finite element analysis, Plane Stress and Plane Strain are two-dimensional simplifications used to reduce complex 3D problems to 2D, significantly decreasing computational effort while maintaining accuracy for specific geometries and loading conditions.
Plane Stress occurs in thin plates or shells where the thickness is small compared to other dimensions and the loading is applied only in the plane of the structure. In this state, the stress components normal to the plane are assumed to be zero:
Applications include thin-walled structures such as thin plates under in-plane loading, sheet metal components, and gear teeth. The strain in the thickness direction is non-zero due to the Poisson effect.
Plane Strain occurs in structures where one dimension is very large compared to the other two, and the cross-section remains constant along that length. The loading is uniform across the length and acts perpendicular to the longitudinal axis. In this state, the strain components normal to the plane are assumed to be zero:
Applications include long structures such as dams, retaining walls, tunnels, and pipes subjected to uniform internal or external pressure. While the strains are zero, the stress in the longitudinal direction is non-zero and is required to maintain the zero-strain condition.
Illustration of Differences:
- Plane Stress: Focuses on thin members where σz = 0. The material is free to deform in the thickness direction.
- Plane Strain: Focuses on long members where εz = 0. The material is constrained from deforming in the longitudinal direction, leading to induced stresses in that direction.
Derive the Quadratic shape function for a bar element.
Answer
A quadratic bar element is a three-node element with nodes at the two ends (x = 0 and x = L) and one node at the mid-point (x = L/2). The displacement field u(x) is represented by a quadratic polynomial:
To derive the shape functions, we define the nodal displacements as u1, u2, and u3 at coordinates x1 = 0, x2 = L, and x3 = L/2 respectively. Applying these boundary conditions:
- At x = 0: u(0) = a0 = u1
- At x = L/2: u(L/2) = a0 + a1(L/2) + a2(L/2)2 = u3
- At x = L: u(L) = a0 + a1L + a2L2 = u2
Solving for constants a0, a1, and a2 in terms of nodal displacements:
Substituting these into the displacement equation u(x) = N1u1 + N2u2 + N3u3 and grouping terms, we obtain the shape functions:
These functions satisfy the properties of shape functions: Ni = 1 at node i and 0 at other nodes, and the sum of shape functions equals unity.
Determine the stresses and reaction for the following truss element. Given E = 70 GPa and A = 500 mm2.
Answer
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Determine the end reactions and deflection at mid span. Assume E = 200 GPa.
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For the axisymmetric body rotating with constant angular velocity ω = 200 rpm, determine the body force vector. Specific density is 7800 kg/m3.
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Derive the jacobian matrix of two dimensional four noded isoparametric element.
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A square cross section of a steel shaft is as shown. Evaluate nodal values, if shaft is 8 mm square. Given G = 8 × 106 N/cm2, and ϕ = 0.005 degree/cm. Use only one fourth of the cross section.
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Evaluate eigen values and eigen vectors for a cantilever beam of length 1 m free at the other end. Take E = 200 GPa, I = 40 × 10-10 m4, A = 2 × 10-4 m2 and weight density = 7850 kg/m3. Use one element method.
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