Consider the pictured bar with axial loads of PB = 50 kip and Pc 50 kip and Pc = 68.5 kip, and the indicated material properties. A B PC. PB C A = 5 in² E 12 x 103 ksi Jy = 35 ksi for this material D PB 18 in PC 24 in 36 in 0.01 in a)For the chosen values of PB and Pc, determine the reaction at D b)Given PB = 50 kip, what value of Pc will produce yielding of the bar? will it secu kip kip
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- Consider the pictured bar with axial loads of PB = 46 kip and Pc = 32 kip, and the indicated material properties. A PB BC AB PB 18 in B PC. PC 24 in A = 5 in² E = 12 x 10³ ksi dy = 35 ksi for this material C D 36 in a)For the chosen values of P_B and P_C, determine the reaction at D 48.0 b)Given PB = 46 kip, what value of Pc will produce yielding of the bar? and in which section will it occur? O CD 0.01 in o kip kip3. A 5-m rod having a uniform cross-sectional area of 1000 sq.mm. is secured between two walls5 meters apart. The load on the rod is zero at 27 °C. E = 200 GPa, α = 11.25x10-6 m/m-°C. Whenthe temperature rises to 57°C:a) What is the stress in the rod assuming that the walls are rigid.b) What is the stress in the rod if the walls spring together a total distance of 0.5 mm.b) A composite bar is shown in figure Qlb. Determine. The thermal stresses in copper tube and steel bar i. ii. Reaction in the steel bar. Copper tube (a1, E1) Steel bar (a2, E2) L = 500mm Fig Qlb Given: di = 100mm, L = 500mm d2 = 200mm, AT = 100°C Соррer Steel E (GPa) 100 200 al°C 16х 10-6 12 x 10-6
- Question 4 Consider the same truss again here as shown below. Suppose that the cross-sectional area of the first element is A1=200 mm?, and all members are made from a same material with Young's modulus of E=55 GPa. Find the stiffness matrix of the first element in N/m. Work out the summation of all terms located on the main diagonal of this stiffness matrix. 100 mm (1) (2) (3) 3 4 150 mm 200 mm Your Answer: Answer Hide hint for Question 4 The summation of all terms located on the main diagonal of matrix т1 т2 M is m1 + m4. т3 т4 100 mm 2.Element 1 Ук ** Answer in MPa. 2 cm Element 2 Assume the reaction forces at the origin were solved to be Rf = [325, 100, 0] N and = Rm [10, -2, 5] Nm. We will consider a hypothetical cut at the connection of the dolly to the axle. Z is positive out of the page. What is the value of normal stress in the x-direction Ox for element 2? Answer options: a) Ox < -300 b) -300 ≤ Ox < -100 c) -100 ≤ Ox < -200 d) 0 ≤ Ox < 100 e) 100 ≤ Ox < 300 f) 300 ≤ OxLet's consider a rod having a solid circular cross-section with diameter of 5 mm and it is made of a material having a Young's modulus E = 200 Gpa and a Poisson's ratio of 0.3. If a tensile force F is subjected to that rod cross-section, the diameter becomes 4.995 mm. determine the applied force F. Select one: O F = 5236 N O F = 6283 N %3D O F= 13090N O F = 10472 N O F= 15708N OF=4189 N 9:22 PM EN ? 4/15/2021
- H.W 1. Consider a two degree of freedom bar elements as shown in figure. Using finite element method to formulate the equilibrium equation of it. If the cross sectional area is 12 mm and E=200 GN/m². 20KN +30KN 15KN 300 * 600- + 350 All Dimension in mm 2. Consider a two degree of freedom bar element as shown in figure. Using finite element method to formulate the equilibrium equation of it, and then estimate the stress distributions. If Esteel-200 GN/m, Ecopper 110GN/m and EAL= 120 GN/m?. d=3 Steel AL Соpper 20KN - 30KN →15KN 300 * 200 - 400 * 350 All Dimension in mmClear 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..1. A circular rod of diameter 20mm and 500mm long is subjected to atensile force 50kN. The modulus elasticity for steel may be taken as 200kN/??2. Find stress and strain due to applied load.
- O P = 287 N OP= 597 N Let's consider a rod having a solid circular cross-section with diameter of 4 mm and it is made of a material having a Young's modulus E = 200 Gpa and a Poisson's ratio of 0.3. If a tensile force F is subjected to that rod cross-section, the diameter becomes 3.995 mm. determine the applied force F. Select one: O F = 13090 N OF= 15708 N OF=4189N F = 5236 N OF= 10472 N O F= 6283N 10:29 EN 4/15/2 TOSHIBA2 Consider a two degree of freedom bar element as shown in figure. Using finite element method to formulate the equilibrium equation of it, and then estimate the stress distributions. If Esteel 200 GN/m², Ecopper= 110GN/m² and EAL= 120 GN/m. d=3 Steel AL- Copper 20KN -30KN →15KN 300 200* 400 - 350 → All Dimension in mm80 mm 170 mm → A В 125 mm |D C 75 mm VF E The frame shown in the provide figure has Load F applied at point D. Answer the following questions with load F = 600 N. Include units in your answer (example Px = 55.4 kN (left)) 2-A. Determine the reaction(s) at the supports at Points A 2-B. Determine the reaction(s) at the supports at Points E 2-C. Extra Credit: Identify any two-force members in this frame