A cross-sectional area of 322.58 mm of an aluminum bar which carries the axial loads applied at the positions shown in figure below. Find the total change in length of the bar if Modulus of elastic (E) = 68947.45 N/mm. Assume the bar is suitably fixed to prevent lateral buckling. 31.13 KN 22.24 KN
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- Rigid bar ACB is supported by an elastic circular strut DC having an outer diameter of 16 in. and inner diameter of 15.3 in. The strut is made of steel with a modulus elasticity of E = 29,000 ksi. Point load P = 4.3 kips is applied at B. 4 ft -5 ft B A 3 ft P Calculate the change in length (in inches) of the circular strut DC. (Use the deformation sign convention.) in. What is the vertical displacement (in inches) of the rigid bar at point B? (Use the statics sign convention.) in.A steel rod with length equal to 12 ft, has a diameter of 0.8 inches in quarter of its length and a diameter of 0.4 inches the rest. If it is subjected to a tensile load of 6,500 lbs, what will be the total deformation of the bar if it has a modulus of elasticity E = 29.5 10^6 psi.For 10 to 12. The compound bar ABCD has a uniform cross-sectional area of 0.25 in². When the axial force P is applied, the length of the bar is reduced by 0.018 in. The moduli of elasticity are 29 x 106 psi for steel and 10 x 10 psi for aluminum. 18 in. 6 in. 18 in. P. P Steel B C Steel D -Aluminum 10. What is p? 11. What is the total elongation of the two steel sections? 12. What is the elongation of the aluminum section?
- Calculate the crippling load for a column using Euler's theory for long columns, whose effective length is 4 m. Take modulus of elasticity as 'E' kN/m² and moment of inertia as 'I' m4. Consider that both ends of the column are fixed. 1. 0.500 T² El kn 2. 0.0255 m² EI KN 3. 0.0625 T² EI KN 4. 0.025 T² EI KNA steel bar with dimensions and cross section shown is suspended vertically. Three concentric downward loads are applied to the bar: 22 250N at the lower end; 13 500 N at 0.35m above the lower end, and 9000N at 1m above the lower end. The modulus of elasticity of the steel is 210GPa. What is the total change in length of the bar?N for Newton, m for meter, mm for millimeter, N/(mm^2) for Stress, mm^2 or m^2 for Area, mm^4 for Moment of inertia and Nm for bending moment. Use brackets if the power is MINUS for Example: 0.00125 N =1.25*10^(-3)N. A member is formed by connecting two steel bars shown below. Assuming that the bars are prevented from buckling sideways, calculate the magnitude force "P" that will cause the total length of the member to decrease 0.56 mm the values of elastic modulus for steel is 215 MPa. Take H1 = 4 cm, H2 = 9 cm, top bar (R) is 3 x 3 cm and bottom bar (Q) is 5 x 5 cm. Also, determine the stress-induced in each section of the bar. Solution: i) Magnitude Force, P = ii) Stress-induced in the top bar = iii) Stress-induced in the bottom bar =
- . A short steel tube of external diameter 70 mm and internal diameter 40 mm is surrounded by a brass tube of same length and having external diameter 90mm and internal diameter 80 mm. The tubes are rigidly fixed and an axial load of 20 kN is placed on the tubes. Find the load carried in each tube and shortening of each tube. Take L = 30 cm, modulus of elasticity of steel is 2.1 x 105 N/mm2and modulus of elasticity of brass is 1.1 x 105 N/mm2.A stepped circular bar having diameters 20 mm, 15: mm and 10 mm over axial lengths of 100 mm, 80mm and 60mm is subjected to an axial tensile force of 5 KN. If E= 100 x 103 N/mm2 and 1/m = 0.32 for the material of bar, determine total change in length and change in each diameter.Clear my choice Let's consider a rod having a solid circular cross-section with diameter of 6 mm and it is made of a material having a Young's modulus E = 120 Gpa and a Poisson's ratio of 0.33. If a tensile force F is subjected to that rod cross-section, the diameter becomes 5.998 mm. determine the applied force F. Select one: O F= 5712 N O F= 2285 N O F= 8568 N O F= 2856 N O F= 3427 N O F=7140 N Finish attempt..