can you please step by step to understand that I did not get it ????? A manufacturer of single-seater racing cars was considering materials andthicknesses of a sandwich panel concept for the chassis. The design case assumeda 600mm wide simply supported panel of the form shown below, subject to a centralpoint load of 1500N. The distance between the supports was 1.20m.D =8 =To a reasonable approximation the expression for panel stiffness, D, is:Estsbd²2:[5.44mm]✓1.20mMaterialSMCCore1500Nand the mid-span deflection, 8, is given by:WL³ WLtc+48D 4bd² Gcwhere:where:Some data for candidate materials are shown in the table below.Young'sModulusEs = Young's modulus of skints= thickness of skinto = core thicknessbd = tc + 2.ts/2 = tc+tsDensity(Mg.m-³)1.80.2tcd h= panel width = 600mmW = 1500N,Span, L = 1.20mShear modulus of core material, Gc= 50MPa(GPa)10i. Calculate the mid-span deflections, of a panel with 40mm core thickness,constructed with 2.0mm thick sheet moulding compound (SMC) skins. ii. Using the data from part i, calculate the overall mass of a 1.2m long x0.6m wide sandwich panel made with SMC skins.[10.944kg]iii. Calculate the maximum bending stress in the skin and the shear stress inthe core for a panel made with SMC skins.[9.35 MPa and 0.03 MPa, respectively]The expression for bending stress, σs, is:Mhbt d²=Oswhere M = applied bending momenth = overall panel thicknessThe expression for shear stress, tc, is:Qbdwhere Q = shear force.Tc=

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A manufacturer of single-seater racing cars was considering materials and thicknesses of a sandwich panel concept for the chassis. The design case assumed a 600mm wide simply supported panel of the form shown below, subject to a central point load of 1500N. The distance between the supports was 1.20m.

A manufacturer of single-seater racing cars was considering materials and thicknesses of a sandwich panel concept for the chassis. The design case assumed a 600mm wide simply supported panel of the form shown below, subject to a central point load of 1500N. The distance between the supports was 1.20m.

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter3: Transient Heat Conduction

Section: Chapter Questions

Problem 3.4P

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E and 5 v X O file:///C:/Users/Hp/Desktop/mm/OUTCOME%20NO.2%20and%205%20with%20Problems.pdf + O Fit to page D Page view A Read aloud 1 Add notes Problems 1. A brass rod of diameter 25 mm and length 250 mm is subjected to a tensile load of 50 kN and the extension of the rod is equal to 0.3 mm. Find the Young's Modulus or Modulus of Elasticity. 99+ to search hp delete prt sc 14 DI backs & 7 8 24 4 P EJR -0 J K H D. F
7. A steel connecting rod is subjected to a completely reversed axial load of 160 kN. Suggest the suitable diameter of the rod using a factor of safety 2. The ultimate tensile strength of the material is 1100 MPa, and yield strength 930 MPa. Neglect column action and the effect of stress concentration. [Ans. 30.4 mm]
Problem 1: (50 points) In Avengers 2, Captain America's improved shield is made from unobtainium, a new material that will soon be available in a store near you. Unobtainium has the normal stress- strain diagram shown. The proportional limit, the elastic limit and the yield point are identical in this material. o [MPa] 2001 175 150 125 100 Fig.1 Normal stress-strain 75 50 25 0 0. 0.05 0.1 0.15 0.2 0.25 Unobtainium has a Poisson's ratio of 0.3. a. [6pts] Identify the yield point oy, the ultimate stress ou and the fracture stress of. Include units. c. [25pts] A bar of unobtainium has a length of 1.5 m, a width of 100mm and a height of 50 mm, as shown. The cross-sectional area is 50mm 100mm = 5 x 10-³m². The bar is subjected to an axial force of 375 kN. Find the normal stress in a cross-section of the bar. b. [5pts] How do you find Young's modulus from this graph? Find its value with units. whesta fins dus al hoss P = 375 KN 1.5 m 0.3 50 mm 100 mm & [mm/mm] Is the material within its…
A house wall is composed of 3 layers in series to each other. There is an Aluminum layer that is 4 cm thick, Copper layer that is 6 cm think and Gold layer that is 3 cm thick. What is the total conductivity of the wall and its R-value in units of ft2.F.hr/Btu? [KAl = 2.37 W/cm.C, KCu = 4.01 W/cm.C, KAu = 3.17 W/cm.C]
During the tensile test, experiment on a sample of mild steel, data represented in the figure below was obtained with an initial diameter (do) of 0.505 inch. At failure, the reduced diameter (d;) of the sample was 0.305 inch. Gauge Length (Lo) is 2 inches and final Gauge Length (L) is 2.625 inches. Answer the questions below: Us TR 70- 60 - Lo LyP *IR 50 - EL PL: proportimal limit EL Elastic limit UXP = 4pper yield point Lyp : Lower us : uiti mated streagth IR : Indicanteed Strengty Rupture TR: True strength Ruptue 40 AL or 8t P 30 **PL (proportimal limt) Direct strain E strain (in/in) aStress Iblinz (esi) 0.003. ogo o-200
Question 1: [Simple stresses] 1. A hydraulic system is used to apply a force in an external fixation device to hold broken ends. The piston of this hydraulic ram is 30 cm in diameter and the water pressure (shaded portion of the sketch) is 3 MN/m² as shown in Figure Q1.1. 3 MN/m? 30 cm 4 cm Figure QI.1 Hydraulic system Determine: 1.1. The stress in the piston rod which has dimeter of 4 cm 1.2. The elongation of the 2 m long piston rod. Take Young's modulus as E = 200 GN/m²
The purpose of this problem is to show the relationship between material constants typically used in engineering practice. This is useful because you may often have access to measurements of or tabulated values of some constants (e.g., Young's modulus and Poisson's ratio) but need another constant (e.g., shear modulus) for a calculation. Use the expression Cijkl = µ(dildjk + dikdjl) + Ad¿jdkl to derive the following: (a) Young's modulus, E = µ(3X+2µ)/(X+μ), from the definition 11 = Ee11 in a unconfined (022 = 0,033 = 0) uniaxial tension test. (b) Poisson's ratio, v = \/(2(X + μ)), from the definition v = €22/11 in the same test as in (a). (c) Shear modulus, μ = G = 012/(2€12) = E/(2(1 + v)). Use the results to show that C can also be written Cijkl Ev/((1+v)(1 − 2v))dij§kl. = E/(2(1 + v))(duðjk + dikdjl) +
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Simple Machines-Levers, Inclined Planes, and Pulleys Directions: Use the appropriate equation to answer the following questions. All answers should be recorded below or in your Engineering Design journal. Remember to show all work. A lever has an effort arm that is 5 meters long and a resistance (load) arm that is 3.5 meters long. How much effort is needed to lift a 100 Newton weight? 5K i). Draw the figure representing the problem ii). How much effort is needed to lift a 100 Newton weight? iii). What is the Actual Mechanical Advantage 1. 5m i) AOON 5-5m
of the Origin. Overall drafting standard - Create the illustrated part. Note the location 1.50 1.25 ANSI. Calculate the overall mass of the illustrated model. Apply the Mass Properties tool. Think about the steps that you would take to build the model. 2.25 Given: A = 3.50 B= 70 Material: 1060 Alloy Density = 0.0975 Ib/in^3 Units: IPS Decimal places = 2 Review the provided information carefully. Units are represented in the IPS (inch, pound, second) system. A = 3.50in, B = .70in bliud of s Origin PAGE 2-95 Eniter insert Cut
۱۰:۱۳ docs.google.com * Fixed-Fixed bar was 20 centigrade degree at room temperature, Determine the thermal stress when the bar is heated to :110 degree A Lo Est = 200 Gpa = 11.7 μm/m°C ας ast Ast = 200 mm² _187.2 MPa (compression) +187.2 MPa (compression) 210.6 MPa (compression) 210.6 MPa (Tension) + 210.6 MPa (compression) + 210.6 MPa (Tension) +187.2 MPa (Tension) O non of these _187.2 MPa (Tension) (
Q: Figure 2 shows a typical mast from a modern racing yacht. It is built in the form of a thin-walled tube (hallow cylinder). To win long-haul ocean races, these yachts need to have every advantage that high-performance materials can give, in terms of maximum stiffness (must not easily deflect), plus minimum weight as possible. The following are the equations to use: 1(F13 8 == 8 EI and I = ar³t, A = 2art assume l and cross section are fixed but thickness t is free to change. • Drive the performance index for this problem • Use table 1 to select top three materials with the best performance index. Figure 2 Table. 1 Data for Tube of Given Stiffness p (Mg.m 3) 2.0 Material E (GPa) GFRP 12 Steel 7.8 200 Wood 0.6 12 Aluminum alloy Titanium alloy 2.7 69 4.5 120