2. A beam with a square cross section of 40 mm x 40 mm is loaded...

 

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2. A beam with a square cross section of 40 mm x 40 mm is loaded in torsion with an angle of twist 8 = 2.0 x 104 mm-¹. Since the cross section, shown in figure Q2, is constant, it can be represented in 2D and the governing equation can be given as a²p(x,y) a²p(x,y) + ax² Əy² where p(x, y) is the stress function and G is the shear modulus. 4 3 and 5 k(e) = 40mm Figure Q2: Geometry and boundary conditions for a beam with a square =-2GB cross-section. Due to symmetry in the beam's cross section, the beam may be represented using the elements and nodes shown in figure Q2. The discrete form of the problem can be given as [K]{4} = (F) The element stiffness matrix and load vector may be given as b³ + c² 1 44(e) b₂b₂ + C₂C₁ b3b₁ + C3C₁ 4A b₁ b₂ + C₂ C3] b₂b3 + C₂C3 b3 + c} Finite element nodes f(e) - b₂b₂ + C₁ C₂ b + c b₂b₂ + C3C₂ 20GA(e) 3 --1(e) where A(e) is the area of an element, G is the shear modulus of the beam material, b₁ = y2-ys, b₂ = y3- Y₁. b3 =Y₁Y2, C₁ X3 X2 C₂ = X1 X3 and c3 = X₂-X₁. Here x₁-3 and y₁-3 are the Cartesian coordinates of the 3 nodes in a linear triangular element. a) Using the elements shown in figure Q2 and given a shear modulus of G = 30,000 N/mm², calculate the distribution of stress function p(x, y) in the beam. Take (x, y) = 0 at the surface of the beam. 2. A beam with a square cross section of 40 mm x 40 mm is loaded in torsion with an angle of twist 8 = 2.0 x 104 mm-¹. Since the cross section, shown in figure Q2, is constant, it can be represented in 2D and the governing equation can be given as a²p(x,y) a²p(x,y) + ax² Əy² where p(x, y) is the stress function and G is the shear modulus. 4 3 and 5 k(e) = 40mm Figure Q2: Geometry and boundary conditions for a beam with a square =-2GB cross-section. Due to symmetry in the beam's cross section, the beam may be represented using the elements and nodes shown in figure Q2. The discrete form of the problem can be given as [K]{4} = (F) The element stiffness matrix and load vector may be given as b³ + c² 1 44(e) b₂b₂ + C₂C₁ b3b₁ + C3C₁ 4A b₁ b₂ + C₂ C3] b₂b3 + C₂C3 b3 + c} Finite element nodes f(e) - b₂b₂ + C₁ C₂ b + c b₂b₂ + C3C₂ 20GA(e) 3 --1(e) where A(e) is the area of an element, G is the shear modulus of the beam material, b₁ = y2-ys, b₂ = y3- Y₁. b3 =Y₁Y2, C₁ X3 X2 C₂ = X1 X3 and c3 = X₂-X₁. Here x₁-3 and y₁-3 are the Cartesian coordinates of the 3 nodes in a linear triangular element. a) Using the elements shown in figure Q2 and given a shear modulus of G = 30,000 N/mm², calculate the distribution of stress function p(x, y) in the beam. Take (x, y) = 0 at the surface of the beam.

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