some value P, and the final deflection in the direction of the load is A, the load does work U, =PA on the structure. By conservation of energy, this work is stored in the elastic deformation energy of the structure, U¡ . For a truss, the internal strain energy is the sum of the axial strain energies in all the elements. This gives the equation N²L PA =E 2AE where N is the internal axial force in each member.

some value P, and the final deflection in the direction of the load is A, the load does work U, =PA on the structure. By conservation of energy, this work is stored in the elastic deformation energy of the structure, U¡ . For a truss, the internal strain energy is the sum of the axial strain energies in all the elements. This gives the equation N²L PA =E 2AE where N is the internal axial force in each member.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

See similar textbooks

Similar questions

Q1 (70 P). Please look at the Figure 1 (a) Please find the equation of the elastic curve f (b) Please find the slope at A, B -2L- L/2- Figure 1
Question 4 Consider a thin rectangular cantilever beam of linear elastic and isotropic material in Figure 2. The solution to this two dimensional plane stress problem under point load, F and moment, M is obtained from the Airy's stress function as: i. ii. Φ y M 40313. F Figure 2 ——▬▬▬▬▬▬▬▬▬▬ y² Determine the stress components. Using boundary conditions, investigate how the load F and M are applied to the beam to satisfy the given Airy function. 4c X OL 2c t=1
An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0x10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 270 mm, a cross-sectional area of A = 13520 mm², and a length of L = 4.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1830 με. Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. P Answer: Δd = i d eTextbook and Media Save for Later L mm Attempts: 0 of 5 used Submit Answer
-450 - 450 –→-350 y"D B 600 18 KN Figure 3 A) Given Autodesk inventor and Ansys software, write one paragraph summary each to describe how you would conduct a simple simulation. B) A truss assemblage shown in Figure 3, in which node A experiences a downward load of 18 kN, lies in the global x-y coordinate. If, A = 2.5x10* m², E = 200 GPa for all elements, use the direct stiffness method to: i) Assemble the global stiffness matrix. ii) Solve for the nodal displacements. ii) Compute the stresses in the elements. 2.
3. Consider a frame as shown below. The vertical section AB is 3 m long, and the horizontal section BC is 1.5 m long. The entire frame has a hollow circular cross- section with outer radius 5 cm and inner radius 4.5 cm. The frame is loaded as shown, with a load Py = 2500 N in the -y-direction (see coordinates below) and a load P, 2500 N in the -z-direction. =\ B ↑ Зм 1.5~ Pz - ♡ ४ P. a. Calculate the stress state at the base of the frame (point A) at each of the points indicated on the figure (a, ß, y, 8). You may neglect stresses you deem to be small, but please state such assumptions clearly. b. Using the results from (a), calculate the maximum tensile stress, compressive stress and shear stress at point A. c. Assuming the material used for the frame fails in tension at 400 MPa and in compression at 500 MPa, calculate the value of P, P, at which failure occurs. Where does failure occur? HE
A combined section of steel and copper is subjected to a compressive load. Refer to the table below for the properties of each Steel Сopper Area, A 900 mm2 1200 mm2 Modulus of elasticity, E 200 GPa 120 GPa Allowable compressive stress 140 МPа 70 MPa Which of the following most nearly gives the load, in percent, carried by the copper? Select the correct response 44. 44% 55.56% 66.67% 33.33%
Rigid bar ABCD is loaded and supported as shown. Steel [E = 30800 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.78 in.2 and bar (2) has a cross-sectional area of 0.32 in.2. After load P is applied, the strain in bar (1) is found to be 610 μεμε. Assume L1=50 in., L2=76 in., a=28 in., b=21 in., and c=22 in. Determine:(a) the stresses in bars (1) and (2).(b) the vertical deflection vD of point D on the rigid bar.(c) the load P.
Q2/ A force P = 1 KN applied to a right bar suspended by three wires, as shown below. all wires are of equal size and the same material. for each wire, A = 80 mm , E =200 GPa , and L =4 m. If initially, there were no slack in the wires, how will the applied load distribute between the wires? 200 100 100 4000 mm mm mm mm
E = 70 Gigapascals (member QR) A = 1000 mm^2 (QR) 500 mm E = 200 Gigapascals (member TV) T A = 1500 mm^2 (TV) V M 60⁰ 75 100 100 mm 200 mm W E = 150 Gigapascals (member UW) A = 2500 mm^2 (UW) As may be seen in the diagram, the DEFORMABLE bars QR, TV, and UW are pinned to the RIGID bar RUST, which is pinned at S. Find the maximum moment M (in KN-mm) that may be applied at S without causing the rigid bar RUST to rotate more than 0.0005 degrees about S.
L1 PB 2 E2 L2 D2 Dimensions: L2 L1 D2 D1 17 mm 18 cm 23 cm 15 mm Problem Statement: The rod shown is subjected to loads of PR = 112 kN and Pe = 18 kN. Segments AB and BC have elastic moduli of E, = 206 GPa and E = 73 GPa. The rod has circular cross-sections, and the dimensions of the rod are given above. Find the overall change in length of the rod under this load. Assume: • The collar at B is rigid • Stress concentration effects may be neglected 2014 Michael Swanbom
A solid bar of length L = 4 m and diameter 200 mm is heated from 20 to 320 degrees celsius and restrained between two solid immovable walls. Young's modulus of the material is 90 GPa and the coefficient of thermal expansion is 22 x 10-6/°C. →Calculate the thermal strain, e, in micro-strain correct to two decimal places. micro-strain. Calculate the thermal stress, o, in megapascals (MPa) correct to two decimal places. E: σ: MPa. Hence calculate the force exerted by the bar in meganewtons (MN) correct to two decimal places: F MN. L
W1 F1 F2 A C 2.0 m L2 1.0 m A beam is supported by a pin connection at B and a rollor support at D. An upwards force Fi =1.5 kN acts at the loft end of the beam (point A). A rightwards force F17.5 kN acts at the right end of the bearn (point E). A unitormly distributed load w1.8kN/m pushes down on the beam between points C andE The separation betwoen A and Bis L =1 m, and the separation between Cand D is La-2 m