A construction worker punched a bamboo wall (the wall has a shearing strengthof 241 MPa). His hand is equivalent to 0.6% of the mass of his own body, andforms a circular shape with a compressive stress limit of 0.31 mPa before itbleeds.Assumptions:Diameter of his hand= 3 inchesMass of worker= 110 poundsFind for:a. thickness in cm of the hole he made on the bamboo wallb. force he exerted to punch the wall

Answered: Image /qna-images/answer/7ba340d1-43d5-4d0c-949c-fb579d6cd8a8.jpg

Answered: A construction worker punched a bamb00…

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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1a. A steel rim, with a mean diameter of 1 m, rotates at 1 000 r/min. Determine the circumferential stress-induced if the density of steel is 7.8 Mg/m³. 1b. A Cylindrical steel pressure vessel of 400 mm diameter has a wall thickness of 20 mm and is subjected to an internal pressure of 4.5 MN/m². (a) Calculate the tangential and longitudinal stresses in the steel. (b) To what value may the internal pressure be increased if the stress in the steel is limited to 120 MN/m²?
The cover of a cylindrical pressure vessel made of cast iron is shown in fig.1 The inner diameter of the cylinder is 500 mm and the internal pressure is limited to 2MPa. The cover is fixed to the cylinder by means of 16 bolts with a nominal diameter of 20 mm. Each bolt is initially tightened with a preload of 20 KN. The bolts are made of steel FeE250 (S yt = 250 N/ mm 2 ). Assume. E for steel = 207 KN/mm 2 ,E for cast iron = 100 KN/mm 2, E for Zinc = 90 KN/mm 2,Determine the factor of safety for bolts considering the effect of the gasket.
Choose the stress to which the size correction factor should be applied in Soderberg's formula. a. Mean stress b. Yield stress c. Variable stress d. Endurance limit stress Clear my choice
This particular problem was already answered but wasn’t able to understand handwriting. Kindly, write legibly. Thank you in advance. Missing value is MO = MT 170 kN.m Solve for: a) Determine the shear stress in semi-circle portion of the profile b) Compute the maximum torque that can be applied if the allowable shear stress is 200 MPa
As information for the mechanic, it is desired to evaluate the variation in shear force along the length of the shaft. Therefore, what is the value of the shear force in the central section of the shaft (125 mm < x < 600 mm)? Options: a) The shear force in the central section of the shaft is 0 kN. B) The shear force in the central section of the shaft is 0.8 kN. c) The shear force in the central section of the shaft is 1.5 kN. d) The shear force in the central section of the shaft is 1.48 kN. e) The shear force in the central section of the shaft is 0.82 kN.
Q.1 A rotary shaft carries distributed load of 250 w/m with a long 3m. The shaft transmits 35 KW at 290 rpm. A shaft is simply supported shaft. The mux. shear stresses 56 MPa and the allowable Compression stress 45 MPa, the safety factor is 1.5. Find the size of the shaft if it subjected to suddenly applied load.
The stress at a point of a machine component are 200MPa tensile and 100MPa compressive. Using Mohr’s Circle method find the intensities of normal, shear and resultant stresses on a plane inclined at an angle of 200 with the tensile stress. Also find the magnitude of shear stress in the component.
3 Q2. For the machine part shown made from AISI 1050 HR, the machine operate at 300°C, the reliability is 95%,the part is chromium plating, with factor of safety of 1.8. Calculate the value of Moment M against infinite life M 30 20 Dimensions in mm, thickness 8mm
Q2(c) The steel alloy shaft having a diameter of 38 mm is used to transmit torque, T as shown in Figure Q2(c). If the allowable shear stress of the shaft is 80kPa, i)Determine the maximum torque T that can be transmitted by the shaft. ii)What would be the maximum torque T' if a 25 mm diameter hole is bored through the shaft? Figure Q2(c.) |Q2 c) (i) T c The torsion formula :7 = J Т (0.019) T (0.019)^4 ; Therefore, the max torque T = x10 Nm 80 2 Q2 c) (ii) For point lying on the outside suurface of the shaft: x10 ="(0.019) * (0.019*-0.0125 ) ; Therefore the max torque T' Nm 80 2 Q2 c) (iii) The shear stress at point located at inner surface of shaft, T'c T = -= J ) (0.0125) = 7 (0.019*-0.01254) T'c kPa 4. 2 2.
"A bolt assembly is experiencing 1000 lb of force. The diameter of the head is 5in, bolt diameter is 3in with bolt head height of 1 in. Find the shear stress in psi."
Two solid circular shafts are connected through a flange bolt coupling consisting of 3 circular bolt rows. The free ends of the shafts are to be subjected by equal torque. Solid Shafts Diameter (mm) Length (m) Shear Modulus (GPa) Shaft A Shaft B 65 66 3 3.8 81 84.6 Flange-bolt coupling Bolt Circle Diameter (mm) Bolt Alowable Shear Stress (MPa) Bolt Diameter Shear Modulus Number of Bolt Row (mm) (GPa) Bolts 1 95 14 77 35 4 2 116 17 97 42 10 3 135 13 93 49 14 A Calculate the torque capacity of the flange-bolt coupling (kN-m). B. Calculate the stresses in both shaft if the torque capacity is applied at both ends (MPa). C. Calculate the total angle of twist (in degrees).
A drilling riser is used to drill a 4km deep hole into the seabed. The riser can be approximated as a hollow steel shaft of 185mm and 160mm outer and inner diameter respectively. If the power exerted is 230kW and the speed of rotation of the drill is 75RPM, calculate the maximum shear stress in the shaft and the twist of one end relative to the other in degrees.(Note the angle of twist of a 4km drill string is going to be >> 360o ) G = 89 kN/mm2

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