EXAMPLE:9.As shown in the figure, there is a gapbetween the aluminum bar and therigid slab that is supported by twocopper bars. At 10°C, A = 0.18 mm.Neglecting the mass of the slab,calculate the stress in each rod whenthe temperature in the assembly isincreased to 95°C. For each copperbar, A= 500 mm2, E = 120 GPa, and a= 16.8 μm/(m °C). For the aluminumbar, A = 400 mm2, E = 70 GPa, and a= 23.1 μm/(m ·°C).Home)(Practice atCopperAluminumCopper750 mm

Given data: T0=10°C∆=0.18 mmT=95°CFor each copper bar:A=500 mm2E=120 GPaα=16.8 μm/m°CFor each…

Answered: As shown in the figure, there is a gap…

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 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 0.66 mm the values of elastic modulus for steel is 207 MPa. Take H1 = 4 cm, H2 = 11 cm, top bar (R) is 3 x 3 cm and bottom bar (Q) is 7 x 7 cm. Also, determine the stress-induced in each section of the bar. i) Magnitude Force, P = (in N) ii) Stress-induced in the top bar = (in N/mm^2) iii) Stress-induced in the bottom bar = (in N/mm^2)
The composite bar shown in figure 1 is fixed to the wall at A and connected to the wall at B by two bolts of 14mm in diameter. If at temperature of 10°C, the composite bar is just fit between the two walls (stress free). Determine the maximum and minimum temperature the system can withstand without exceeding any of the allowable stresses. Allowable compressive stress of steel =180 MPa Allowable compressive stress of aluminum = 120 MPa Allowable tensile stress of the bolt = 200 MPa Allowable shear stress of the bolt = 120 MPa B A 14 mmedia. bolt steel 14 mne dia- bolt Bolt Aluminura L= 200 mul L = 150 mm Tall 200 MPa E= 73G Pa E,= 200GPQ tension A = 200 mus? 'all shear = 1 20 HP A 400 mana K= 23 x10/ x, = 12x10/d al 180 MPa all = 120 MPa Comp Comp Figure 1
A 2.5m steel rod with a cross sectional area of 1250m^2 is secured between rigid supports at both ends. If there is no stress of 30°C, E = 200000MPa. (a)Compute the stress when the temperature drops to 0°C, if it has a coefficient of thermal expansion of 0.0000117 m/m°C. (b)Compute the strain of the rod if the rod supports yield by 0.800mm as temperature drops. (c)Compute the stress in the rod if the support yield and moves together a distance of 0.800mm as the temperature drops.
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
A circular bar of length 400 mm and tapering uniformly from 50 mm to 25 mm diameter is held between rigid supports at the ends. Calculate the maximum and minimum stress developed in the bar when the temperature is raised by 30°C. Take E = 2 × 105 N/mm² and a = 1.2 x 10-5. per °c
Figure below shows the bar with three equal elements. Use the finite element method and calculate: 2.1 the global stiffness matrix 2.2 the displacement on node 2, 3, 4 2.3 the Strain in each element 2.4 the stresses in each element using Hook's law and compare with theoretical stresses (o=F/A) (1) (2) (3) -50 N 10 mm 10 mm 10 mm A1=50 mm? A2=20 mm? A:=10 mm? E=200 GPa
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Two round solid bars of different materials are connected in series against rigid supports as shown in Figure 4. Determine the stress in each bar and the support reactions if temperature is increased by 18° C. Bar A da, EA, A Bar B dв, Ев, ав LA LB Figure 4 Use the following information. Bar, B: Length, LB Diameter, dB Poisson's Ratio Young's Modulus EB = 210 kN/mm2 Coefficient of expansion aB = 12.4x106 °C- Bar, A: Length, LA Diameter, dA = 25mm =D40 mm = 10 mm 30 mm S. = 0.28 =D0.32 = 70 kN/mm² Coefficient of expansion aa = 16.5x10-6 °C-I Poisson's Ratio %3D Young's Modulus EA
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
The temperature of an aluminum bar and a steel bar was raised from 40°C to 160°C. If there was a gap 0.4 mm between the two bars at 40°C as show in figure below, determine the normal stress in the aluminum bar and the change in length of the aluminum bar. 0.5 mm 300 mm 250 mm B Aluminum Stainless steel A = 2000 mm² E = 75 GPa A = 800 mm² E = 190 GPa a = 17.3 x 10-6/°C %3D %3D a = 23 X 10-6/°C

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