A overhang beam hanged at B(6m) has two loads of 8 & 4 kn at 6 and 9 m. Calulate its moment using castigliano's theorem. Also calculate Strain Energy.
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- A weight W = 20 kN falls through a height h = 1,0 mm onto the midpoint or a simple beam of length L = 3 m (see figure). The beam is made of wood with square cross section (dimension don each side) and E = 12 GPa. If the allowable bending stress in the wood is °aLLow =10MPa, what is the minimum required dimensionThe wood joists supporting a plank Floor (see figure) are 38 mm × 220 mm in cross section (actual dimensions) and have a span length of L = 4.0 m. The floor load is 5.0 kPa, which includes the weight of the joists and the floor. (a) Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 14 M Pa. (Assume that each joist may be represented as a simple beam carrying a uniform load.) (b) If spacing s = 406 mm, what is the required depth ft of the joist? Assume all other variables remain unchanged.A beam ABC is fixed at end A and supported by beam DE at point B (sec figure). Both beams have the same cross section and are made of the same material. Determine all reactions due to the load P. What is the numerically largest bending moment in cither beam?
- Solve the preceding problem for a wide-flange beam with h = 404 mm, b = 140 mm, bf= 11.2 mm, and rf. = 6.99 mm.-1 through 5.10-6 A wide-flange beam (see figure) is subjected to a shear force V. Using the dimensions of the cross section, calculate the moment of inertia and then determine the following quantities: The maximum shear stress tinixin the web. The minimum shear stress rmin in the web. The average shear stress raver (obtained by dividing the shear force by the area of the web) and the ratio i^/t^ The shear force carried in the web and the ratio V^tV. Note: Disregard the fillets at the junctions of the web and flanges and determine all quantities, including the moment of inertia, by considering the cross section to consist of three rectangles. 5.10-5 Wide-flange shape, W 18 x 71 (sec Table F-l, Appendix F); V = 21 k.Derive the formula for the bending stress of a beam with a rectangular cross section and triangular cross secrion. Thank you
- Give me right solution according to the question. Draw the shear force diagram and Bending moment diagrams for the beam clearly.. Take M= 60 KN-m P=30 KNQ-5. Calculate the maximum allowable internal bending moment M that can be applied to the beam. The cross-section of a steel I beam is as shown in the figure. The allowable tensile and compressive stress values are oallowable tensile =165 MPa and %3D compressive dallowable = 150 MPa . 4 cm 30 cm 30 cm M 4 cm 4 cmA rectangular bar having width twice the depth is used as a beam. The beam is made of mild steel material having elastic modulus of 2.1 x 105 N/mm? and it undergoes bending by external load which makes radius of curvature of 150 m. If the allowable bending stress in the beam is to be limited to 100 MN/m. find the cross section of the beam.
- A cantilever beam of cross-section 90 mm. width 120 mm deep carries a UDL of 12 KN/m. over the entire length and a concentrated load of 15 KN at the right end. Find the bending stress in the beam, when the length of beam is 10 m.Calculate the maximum bending stress for beam shown below, if P = 47 kN, d = 80 mm and b = 200 mm.A beam has a bending moment of 3.5 kN-m applied to a section with a hollow circular cross-section of external diameter 3.7 cm and internal diameter 2.2 cm . The modulus of elasticity for the material is 210 x 109 N/m2. Calculate the radius of curvature and maximum bending stress. Also, calculate the stress at the point at 0.6 cm from the neutral axis (i) The moment of inertia = ii) The radius of curvature is (iii) The maximum bending stress is iv) The bending stress at the point 0.6 cm from the neutral axis is Answer and unit for part 4