The wedge shaped bar shown in Figure Q1 is to be analysed using a single one dimensional quadratic element as shown. The bar has a rectangular cross section with a constant width and a linearly varying depth. un 1 k= 10E-25 211 He 500 + 712' 2 100 a) Calculate the strain shape function matrix, [B], for the element. b) Using the strain shape function matrix derived in (a), show that the stiffness matrix for the element is given by the equation: I 1. Ֆ. Մ. 3 Figure Q1. 112 h -25 5+1 (3- d M 100- ALL DIMENSIONS IN mm with units of N/mm of the material 10 Where: E = Young's modulus c) Show how the strain shape function matrix may be used to determine the stress from the nodal displacement for this type of element. Do not attempt to calculate the stress.

The wedge shaped bar shown in Figure Q1 is to be analysed using a single one dimensional quadratic element as shown. The bar has a rectangular cross section with a constant width and a linearly varying depth. un 1 k= 10E-25 211 He 500 + 712' 2 100 a) Calculate the strain shape function matrix, [B], for the element. b) Using the strain shape function matrix derived in (a), show that the stiffness matrix for the element is given by the equation: I 1. Ֆ. Մ. 3 Figure Q1. 112 h -25 5+1 (3- d M 100- ALL DIMENSIONS IN mm with units of N/mm of the material 10 Where: E = Young's modulus c) Show how the strain shape function matrix may be used to determine the stress from the nodal displacement for this type of element. Do not attempt to calculate the stress.

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

Calculate the stiffness matrix for the Q8 element shown below. Make use of the shape functions in natural coordinates and the Jacobian calculated in previous problems. Use 2x2 Gaussian numerical integration. Use plane strain conditions (for unit thickness) with E=200E9 (MPa) and nu=0.25. I have attached my matlab code below. Please look thru it and correct it. Thanks. u=sym('u') %%ztav=sym('v') %%eta x1=1x2=2x3=3.5x4=3.6x5=4.0x6=2.3x7=1.2x8=0.8 y1=1y2=1y3=1.3y4=2.3y5=3.3y6=3.2y7=3.0y8=2.0X=[x1;x2;x3;x4;x5;x6;x7;x8]Y=[y1;y2;y3;y4;y5;y6;y7;y8]%% diff of ztaDN1Du=1/4*(1-v)*(2*u+v);DN2Du=-u*(1-v);DN3Du=1/4*(1-v)*(2*u-v);DN4Du=1/2*(1-v^2);DN5Du=1/4*(1+v)*(2*u+v);DN6Du=-u*(1+v);DN7Du=1/4*(1+v)*(2*u-v);DN8Du=-1/2*(1-v^2); %% diff of etaDN1Dv=1/4*(1-u)*(2*v+u);DN2Dv=-1/2*(1-u^2);DN3Dv=1/4*(1+u)*(2*v-u);DN4Dv=-v*(1+u);DN5Dv=1/4*(1+u)*(2*v+u);DN6Dv=1/2*(1-u^2);DN7Dv=1/4*(1-u)*(2*v+u);DN8Dv=-v*(1-u);%% find dx/du dx/dv, dy/du and dy/dva=[DN1Du, DN2Du, DN3Du, DN4Du, DN5Du, DN6Du,DN7Du,…
Material 1- BRASS The below shown graph is drawn from the tabulated values of steel which we measured during the experiment of thermal (Diameter = 25mm) conductivity: (Consider the value of heater power (Q') and the area of cross section (A) of the material from the tabulated values) Power X-axis 1 unit =1 cm y-axis 1 unit = 10°C Temperature (°C) 50- Q' 1 2 3 4 5 6 7 8 (W) 14.65 79 77.4 76 50.1 46.5 42.4 36 34.5 : 20- 10- Distance X in cm 123 10 Material 2 - STEEL Calculate the following: Thermal conductivity of (Diameter = 25mm) steel Calculated Value quantities Power Difference in temperature (AT) C° (with sign) Difference in distance Temperature (°C) between test points (Ax) cm Slope from the graph °C/m (with sign) Q' 1 (W) 2 3 7 8 Thermal conductivity of steel ksb W/m°C 14.2 88.6 87.5 85 33.9 33.6 32.4
When stress is applied to a liquid over an application time tp the liquid needs a relaxation time tc to accordingly deform. Relaxation time tc is a physical property of liquids. The Deborah number is De= tdt, . The lower the Deborah number of a stress applied to a material the more.. : Select one: O The material behaves like a solid O The material behaves like visco-elastic O The material behaves like a liquid
Question 2 Study the table below and answer the questions that follow. Coercivity (kA/m) BH max (kJ/m³) Material Curie temperature (°C) A 19.90 8 887 В 0.08 0.04 771 C 1 500 175 723 238.85 28 429 2.2. a) On the same set of axes draw and clearly label the hysteresis loops for materials B and C. b) Explain any two features of the hysteresis loops you drew in (a). c) Suggest an application for material B. Justify your answer.
Fit a power law model to the rheological behavior presented in the data: T [=]D/cm^2 y[=]1/2 1 3 2 - 3 30 4 52 5 80 6 100 7 130 8 160 9 175 10 190 11 218 12 240 13 265 (The second data in the table is not necessary to solve it.)Help yourself with matlab to solve it with the following formula: (Photo)
When a beam section is subjected to a shear load, a shear stress distribution is developed on the section. The distribution of the shear stress is not linear. Elasticity theory can be used to calculate the shear stress at any point. However, a simpler method can be used to calculate the average shear stress across the width of the section, a distance y above or below the neutral axis. The average VQ shear stress is given by T = Here V is the shear It force on the section, I is the moment of inertia of the entire section about the neutral axis, and t is the width of the section at the distance y where the shear stress is being calculated. Q is the product of the area of the section above (or below) y and the distance from the neutral axis to the centroid of that area (Figure 1). In short, Q is the moment of the area about the neutral axis. I Figure 1 of 1 An I-beam has a flange width b = 250 mm, height h = 250 mm, web thickness tw = 9 mm, and flange thickness tf = 14 mm. Use the…
Q4. Two are applied to the pipe AB as shown. Knowing that the pipe has inner and outer diameters equal to 35 and 42 mm, respectively. Determine the normal and shear stresses at (a) point a, (b) point b.
Q3/ In real life engineering application, it is required to design a spring in train, which one of the following materials in the table below should be choose, justify your answer ? used correct units? Poisson's Yield Strength Modulus of Elasticity 106 psi Material MPа psi GPa Ratio 30 0.30 120000 207 Steel alloy Brass alloy Aluminum alloy Titanium alloy 830 0.34 0.33 380 55000 97 14 275 40000 69 10 690 100000 107 15.5 0.34
Find the axial stress (Subscript[\[Sigma], z]) and axial strain (Subscript[\[Epsilon], z]) for a thick-walled circular pipe subjected to an internal pressure Subscript[p, i] for the following three cases:a) the pipe is long and open (plane strain)b) the pipe is short and open (plane stress)c) the pipe is is closedExpress the above solutions in terms of the pressure Subscript[p, i], the the pipe inner and outer radii and the material constants.
Temperature (°C) 1600 0 1500 1400 1300 1200 1100 1000 1085°C 0 (Cu) 1 20 20 Alloy Brass Steel Composition (at% Ni) 40 Liquid Liquidus line Aluminum Titanium B @x+L 60 CE 40 60 Composition (wt% Ni) 80 Yield Strength [MPa (ksi)] 345 (50) 690 (100) 275 (40) 480 (70) Solidus line 1453°C 80 100 2800 2600 Problem 5 A structural member 250 mm (10 in.) long must be able to support a load of 44,400 N (10,000 lb,) without experiencing any plas- tic deformation. Given the following data for brass, steel, aluminum, and titanium, rank them from least to greatest weight in accordance with these criteria. 2400 8.5 7.9 2.7 4.5 2200 2000 100 (Ni) Density (g/cm³) Temperature (°F)
A rectangular tube of outer width w = 7.5 m, outer height h = 5 m and thickness t = 0.17 m experiences bending through the x axis. What is the moment of inertia of the cross-sectional area In? y -- X MATLAB input variables: format shortEng W = 7.5; h = 5; t = 0.17; moment of inertia I, = m4
Calculate the stiffness matrix for the Q8 element shown below. Make use of the shape functions in natural coordinates and the Jacobian calculated in previous problems. Use 2x2 Gaussian numerical integration. Use plane strain conditions (for unit thickness) with E=200E9 (MPa) and nu=0.25. The final answers is F1= 0.18452e12. I have attached my matlab code below. Please look thru it and correct it. Thanks. u=sym('u') %%ztav=sym('v') %%eta x1=1x2=2x3=3.5x4=3.6x5=4.0x6=2.3x7=1.2x8=0.8 y1=1y2=1y3=1.3y4=2.3y5=3.3y6=3.2y7=3.0y8=2.0X=[x1;x2;x3;x4;x5;x6;x7;x8]Y=[y1;y2;y3;y4;y5;y6;y7;y8]%% diff of ztaDN1Du=1/4*(1-v)*(2*u+v);DN2Du=-u*(1-v);DN3Du=1/4*(1-v)*(2*u-v);DN4Du=1/2*(1-v^2);DN5Du=1/4*(1+v)*(2*u+v);DN6Du=-u*(1+v);DN7Du=1/4*(1+v)*(2*u-v);DN8Du=-1/2*(1-v^2); %% diff of etaDN1Dv=1/4*(1-u)*(2*v+u);DN2Dv=-1/2*(1-u^2);DN3Dv=1/4*(1+u)*(2*v-u);DN4Dv=-v*(1+u);DN5Dv=1/4*(1+u)*(2*v+u);DN6Dv=1/2*(1-u^2);DN7Dv=1/4*(1-u)*(2*v+u);DN8Dv=-v*(1-u);%% find dx/du dx/dv, dy/du and dy/dva=[DN1Du, DN2Du, DN3Du,…