2.4-1 The assembly shown in the figure consists of a brasscore (diameter di = 0.25 in.) surrounded by a steel shell(inner diameter dz = 0.28 in., outer diameter dz = 0.35 in.).A load P compresses the core and shell, which have lengthL = 4.0 in. The moduli of elasticity of the brass and steelare E, = 15 x 10° psi and E, = 30 x 10°psi, respectively.(a) What load P will compress the assembly by 0.003in.?(b) If the allowable stress in the steel is 22 ksi and theallowable stress in the brass is 16 ksi, what is the allowablecompressive load Pallow? (Suggestion: Use the equationsderived in Example 2-5.)Steel shellBrass coredi-dsPROB. 2.4-1

given data:-diameter of brass, d1=0.25in=0.00635m⇒Cross-sectional area of brass,…

2.4-1 The assembly shown in the figure consists of a brass core (diameter d, = 0.25 in.) surrounded by a steel shell (inner diameter dz = 0.28 in., outer diameter dz = 0.35 in.). A load P compresses the core and shell, which have length L = 4.0 in. The moduli of elasticity of the brass and steel are E, = 15 x 10° psi and E, = 30 × 10ª psi, respectively. (a) What load P will compress the assembly by 0.003 in.? (b) If the allowable stress in the steel is 22 ksi and the allowable stress in the brass is 16 ksi, what is the allowable compressive load Pallow? (Suggestion: Use the equations derived in Example 2-5.) Steel shell Brass core L. -dz PROB. 2.4-1

2.4-1 The assembly shown in the figure consists of a brass core (diameter d, = 0.25 in.) surrounded by a steel shell (inner diameter dz = 0.28 in., outer diameter dz = 0.35 in.). A load P compresses the core and shell, which have length L = 4.0 in. The moduli of elasticity of the brass and steel are E, = 15 x 10° psi and E, = 30 × 10ª psi, respectively. (a) What load P will compress the assembly by 0.003 in.? (b) If the allowable stress in the steel is 22 ksi and the allowable stress in the brass is 16 ksi, what is the allowable compressive load Pallow? (Suggestion: Use the equations derived in Example 2-5.) Steel shell Brass core L. -dz PROB. 2.4-1

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

Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. The diameter of aluminum rod (1) is 2.50 in., the diameter of brass rod (2) is 1.75 in., and the diameter of steel rod (3) is 3.00 in. Determine the axial normal stress in each of the three rods. Assume P = 5 kips, Q = 3 kips, R = 13 kips and 5 = 24 kips. The normal stress is positive if tensile and negative if compressive. σ₁ = Answers: 0₂ = (1) 03 = (2) i i A Q B C D ksi ksi ksi
A solid steel bar of diameter di = 1.4 in. is enclosed by a steel tube of outer diameter d3 = 2.25 in. and inner diameter d2 = 1.9 in. (see the figure). Both bar and tube are held rigidly by a support at end A and joined securely to a rigid plate at end B. The composite bar, which has length L = 26 in., is twisted by a torque T = 5,100 lb-in. acting on the end plate. Tube A Bar End plate (a) Determine the maximum shear stresses T1 and T2 in the bar and tube, respectively. Round your answers to the nearest whole number. T1 = psi T2 = psi (b) Determine the angle of rotation (in degrees) of the end plate, assuming that the shear modulus of the steel is G = 11.5 × 10° psi. Round your answer to three decimal places. (c) Determine the torsional stiffness kT of the composite bar. Round your answer to the nearest whole. [Hint: Use equations below to find the torques in the bar and tube.] T = T GIp, + G2IP, kT = k-in.
3. The bar shown below is composed of two pieces, AC and CD. The bar is attached to the wall at point A. The length L = 10 in. The cross sectional areas are: AAC = 4 in? and AcD = 2 in?. The Young's Moduli are: EAC =20 Msi and Ecd= 15 Msi Two forces are applied at points B and D: PB = 50 kips and PD = 20 kips. Determine: (a) the reaction force at A (specify –→ or +) and (b) the deflection of point D (specify → or +). (Hint for (b): Think about how many pieces you must subdivide the rod into.) Ans: RA = 30 kips (+), dp = 7.9167×10-3 in (E) A В C D Pp L 2L L
The assembly consists of two-rod brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rods EF and FG of diameter 40 mm, and a rigid cap H as shown in Figure 2. Neglect the size of the collar at F. Assume that the modulus of elasticity Esteel = 200 GPa and Ebrass = 101 GPa. If the force, P = 5 kN, (a) If the horizontal displacement of end G of rod FG is 0.6 mm, determine the new magnitude of P.
2-) Consider the thin (steel) plate in figure given below. The plate has a uniform thickness t= 1 cm., Young's modulus E = 25 x 10° N/cm?, and weight density p= 0,2448 N/cm³. In addition to its self-weight, the plate is subjected to a point load P = 200 N at its midpoint. (Please be careful for the units.) (a) Model the plate with two finite element. 12 cm (b) Write down expressions for the element stiffness matrices and 18 cm element body force vectors. (c) Assemble the structural stiffness matrix K and global load 36 cm vector F.(' P =200N (d) Using the elimination approach, solve for the global displacement vector Q. (e) Evaluate the stresses in each element. ( -. X 6 ст
Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in the figure. One load of P=33P=33 kipskips is applied to the assembly at A, two loads of Q=30Q=30 kipskips are applied at B, and two loads of R=35R=35 kipskips are applied at C. The normal stress magnitude in aluminum rod (1) must be limited to 1717 ksiksi. The normal stress magnitude in steel rod (2) must be limited to 3030 ksiksi. The normal stress magnitude in brass rod (3) must be limited to 2121 ksiksi. Determine the minimum diameter required for each of the three rods.
The figure below shows a section of a column subjected to load P, which is applied 50 mm from the midline of the section's web. The column is made from material, which has an allowable normal stress of 180 MPa (both in tension and compression). The area of the section is 10050 mm?; the principal moments of inertia Imar = 146.554x106 mm and Imin 20.076x10° mm*. !! Determine the maximum allowable load P. 15 mm 150 mm 15 mm 150 mm 50 mm 15 mm 100 mm 100 mm 1194.6 kN 363.6 kN 1809.0 kN O 516.4 kN O 189.5 kN
A vertical load P is applied at the center B of the upper section of a homogeneous frustum of a circular cone of length L, minimum radius rB, and maximum radius rA, as shown in the figure below. It is made of a material with a modulus of elasticity E. Neglect the effect of its weight. Use the following values: rA = 2 in, rB = 0.5 in, L = 3 ft, E = 30 (103 ) ksi, and P = 1800 lbf. Discretize the original geometry into three equal-length finite elements and (a) Compute all the nodal displacements. (b) Compute all the element axial stresses. (c) Plot the displacement field from the finite element method solution. (d) Plot the axial stress field from the finite element method solution.
Consider the two bars of identical material with their elastic modulus E as shown in Figure below. Both bars have the same circular cross-section with diameter of d. The length of the longer bar is 2L and the length of the shorter bar is L. The angle between the bars is B. The system is subjected to a vertical force acting at point Pas shown in the figure with a magnitude of F. Calculate horizontal and vertical components of the displacement of point P. Use engineering notation (ENG display mode on scientific calculators) with 3 significant figures after the decimal point. E (psi) 1.50E+06 d (in) 1.25 L (in) 12 Beta (degrees) 30 F (Ibf) 110 F P L 2L
The assembly consists of two-rod brass C83400 copper alloy rods AB and CD of diameter 30 mm, a stainless 304 steel alloy rods EF and FG of diameter 40 mm, and a rigid cap H as shown in Figure 2. Neglect the size of the collar at F. Assume that the modulus of elasticity Esteel = 200 GPa and Ebrass = 101 GPa. If the force, P = 5 kN, (a) Sketch the free body diagram and calculate the internal forces acting in each section of the rod. (b) Determine the average normal stress developed in rods FG, EF and AB. (c) Determine the average normal strain in the rod of AB.(d) If the horizontal displacement of end G of rod FG is 0.6 mm, determine the new magnitude of P.
Two cylindrical copper alloy bars have the same length L, but different shapes, as can be seen in the figure. The two bars are subjected to the same axial load P. Use the following numerical data: P = 1400 kN, L = 5 m, d = 80 mm, E = 110 GPa, and ν = 0.33. (a) Find the change in length of each bar. (b) Find the change in volume of each bar.
A steel torsion bar AB with diameter d = 34 mm, and ductile material of tensile yield strength Sy = 276 MPa is attached to a rigid arm at A, supported by a bearing at C, and fixed at B, as the picture shows. At the right end of the arm is mounted the wheel on which the vertical force P acts from the ground and which passes through the center of the wheel. If bearing C acts as a simple support and the effect of transverse shear due to P is negligible, determine the maximum force P (in kN) that can be applied to the wheel. Consider a factor of safety n = 2.5. Use the maximum energy of distortion failure criterion and do the analysis on bar AB (torsion bar) in the section where the bearing is.