B. A plastic bar ACB having two different solid circular cross sections is heldbetween rigid supports as shown in the figure. The diameters in the left andright parts are 50 mm and 75 mm, respectively. The corresponding lengths are225 mm and 300 mm. Also, the modulus of elasticity E is 6.0 GPa, and thecoefficient of thermal expansion is 100 x10/Cº. The bar is subjected to auniform temperature increase of 30°C.Calculate the following quantities:(a) the compressive force P in the bar.(b) the maximum compressive stress.(c) the displacement of point C.|75 mmB.|A 50 mm C-225 mm-300 mm

Given, Diameter of Ac bar,d1=50Length of Ac bar, L1=225mmDiameter of CB bar,d2=75mmLength of the CB…

Answered: B. A plastic bar ACB having two…

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

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A steel bar of 30 mm * 30 mm square cross section is subjected to an axial compressive load of 300KN. If the length of the bar is 3m and E=200GPa find contraction of the bar.
3) The rigid horizontal bar AB is supported by three vertical rods as shown in the figure and carries the load of 110 kN. The system is stress free before the load is applied. After the load was applied, the temperature of all wires is raised by 20°C. Use a = 12x10-6/°C for steel, a = 18x10-6/°C for copper and a = 20x10-6/°C_for brass. Neglecting the weight of the bar AB, determine the location of the 110 kN force in order that AB remains horizontal. copper bar: A=195mm? steel bar: brass bar: A=130mm? A=325mm A D 110 kN Note: copper=110 GPA Brass=97 GPA Steel= 200 GPA 3.00m 2.00m 1.80m 1.20m 1.50m
Problem 3: (M4) Your boss at a construction company asked you to verify the thickness of a 3 m long hollow-square steel column (E=200 GPA) that can withstand an axial compressive load of 50 MN, the outer dimension a is 400 mm. The column's cross section is shown in Figure.4. Assume the column is a fixed-free type. Figure.4: Cross-section of a hollow-square steel column
i need the answer quickly
Consider a one-element triangular mesh shown in Figure 2. The edge AC is fixed. The edge BC is subject to lin- early varying traction normal to the edge as shown in Figure 2. Assume plane-stress conditions with thickness 1 = 1 cm and Young's modulus E = 70 GPa and Poisson's ratio v = 0.3. 150 kN/m 50 kN/m C (0,0.3) B (0.4,0.3) a) Specify the boundary condi- tions. b) Construct the stiffness matrix. A (0,0) c) Calculate the global force ma- Figure 2: Triangular element. All coordinates are in meters. trix. d) Solve for the unknown displace- ments and compute the stress at (0.1, 0.2).
Problem 3: (M4) Your boss at a construction company asked you to verify the thickness of a 3 m long hollow-square steel column (E=200 GPA) that can withstand an axial compressive load of 50 MN, the outer dimension a is 400 mm. The column's cross section is shown in Figure.4. Assume the column is a fixed-free type. b Figure.4: Cross-section of a hollow-square steel column
PLEASE ANSWER ASAP TY! The two bars (1) and (2) are made of the aluminum with E = 70 GPa, has the same cross section of 100 mm x 20 mm, and supported by rigid supports at A and C. The coefficient of thermal expansion of the material is 23 x 10–6/oC. At T = 25oC, the two bars have a gap of 5 mm as shown. a) What is the reaction force (in KN) at fixed support A when the temperature is 100oC? b) What is the reaction force (in KN) at fixed support C when the temperature is 100oC? c) What is the stress (in MPa) in bar (1) when the temperature is 100oC? d) What is the stress (in MPa) in bar (2) when the temperature is 100oC?
PROBLEM 2 The composite bar is rigidly attached to unyielding supports. The bar is stress-free at 65°F. At what temperature will the aluminium and steel segments in the given figure have stresses of equal magnitude after the 12.5 kip force is applied? The coefficient of linear expansion of aluminium is a = 12.8 x 10-/°F and steel is a = 6.5 x 10-/ºF Aluminum Steel A = 2 in.2 E = 10 × 106 psi A = 3 in.2 E = 29 × 106 psi 12.5 15 in.- -10 in.-
None

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