Elements Of Electromagnetics
7th Edition
ISBN: 9780190698614
Author: Sadiku, Matthew N. O.
Publisher: Oxford University Press
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- A high-impulse universal testing machine is being used to determine the Young's Modulus of a rectangular solid specimen (l = 20 cm, t = 1.5 cm, w = 2 cm). The machine produces 35,000 Ns that causes the impact force of compression resulted to a 1 mm deformation in the length of the specimen in 0.05 s. Determine the Young's modulus of the specimen expressed in MPa. Round off your answer to the nearest whole numberarrow_forwardDetailed stepsarrow_forwardProblem F2 stress (ksi) 130 120 110 100 90 80 70 60 50 40 30 20 10 0 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 strain (in/in) Above you will find the experimental stress-strain diagram of 1045 steel. Calculate the permanent set if a cylindrical specimen with a diameter of 2 in is loaded to the ultimate stress and then unloaded. Provide your answer in units of in/in with 3 significant figures after the decimal. The elastic modulus of 1045 steel is 29,000 ksi.arrow_forward
- 8arrow_forwardAn alloy steel specimen of 30 mm diameter with a gauge length of 200 mm is load tested. It has an exten- sion of 0.16 mm under a load of 85 kN. Load at elastic limit is 160 kN. Young's modulus of the specimen in GN/m2 isarrow_forward1. Find the strain that results from a tensile force of 1000 N applied to a 10 m long aluminium bar having cross-sectional area of 4x10 m2. The modulus of elasticity of aluminium is 69 GN/m². 2 A resistance wire strain gauge having a nominal resistance of 350 2 is subjected to a strain of 500 micro-strain. Find the change in the value of resistance neglecting the piezoresistive effect.arrow_forward
- Engineering and true strainarrow_forward(d) Define Modulus of Resilience 2. Draw the stress-strain diagram for a typical ductile material and locate the salient points onit 3. A bar of 25 mm diameter is tested in tension. It is observed that when a load of 60 kN is applied, the measured over a gauge length of 200 mm is 0.12 mm and contraction in diameter is 0.0045 mm. Find Poissa 4. A circular rod of diameter 16 mm and 500 mm long is subjected to a tensile force 40 kN. The modulus of e steel may be taken as 200 kN/mm?. Find the elongation and strainin the bar due to applied load. 5. Define (a) Resilience (b) Proof resilience PART -B structions: Part B consists of 3 Units. Answer any one full question from each unit. Each questioncarries 8 marks and may have su 1. A steel circular bar has three segments as shown in the fig. Determine the total elongation ofthe bar. Take E = 210 GPa. 300 kNI 330 kN 30 mm 50KN 15 mm 20 mm 80 kN A + 150 mm 200 mm 250 mm B. D (OR)arrow_forwardA tensile test was conducted on a mild steel& the following data was obtained as follows. Diameter of the steel bar = 3 cm,Gauge length of the bar = 20 cm. Load at elastic limit = 250 kN . Extension at a load of 150kN = 0.21 mm . Maximum load = 380 kN . Total extension = 60 mm . Diameter of the rod at failure = 2. 25 cm Determine: (a) Young’s modulus (b) stress at elastic limit (c)the percentage of elongation (d) Percentage decrease in area.arrow_forward
- The error in a measurement due to strain when using a tape measure depends on the material that it is made of. For a tape with a cross section of 10 mm by 0.7 mm, calculate how much error is introduced in inches into the measurement at a 300 ft length if the tape is pulled with 80 N and if the tape is made of: a) mild steel (E = 30 x 10^6 psi), b) polycrystalline aluminum (E = 10 x 10^6 psi) c) fiberglass composite (E = 3.5 x 10^6 psi in the long direction)arrow_forwardA mild steel rod of 12mm diameter was tested for tensile strength with the gauge length of 60mm. After the test final length and the final diameter were 80mm and 7mm, respectively. Find the strain in the transverse directionarrow_forwardFor a given homogeneous, isotropic, linearly elastic material, E = 15e6 psi and v = 0.3. Solve for the shear modulus. 2.1.1 Homogeneous, isotropic, linearly elastic materials For specimens undergoing small deformations, the stress-strain diagram often ex- hibits a linear behavior. Although this is a very crude approximation to the behavior of actual materials, it is a convenient assumption that is often used for preliminary evaluation. A linear relationship between stress and strain can be expressed as 01 = E €1, (2.1) where the coefficient of proportionality, E, is called Young's modulus or modulus of elasticity. Since strains are non-dimensional quantities, this coefficient has the same units as stress quantities, i.e., Pa. This linear relationship is known as Hooke's law. The elongation of a bar in the direction of the applied stress is accompanied by a lateral contraction that is also proportional to the applied stress. The resulting defor- mations for this uniaxial state of stress…arrow_forward
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