A steel compression member of 9-ft effective length supports an eccentric load as shown. Using the allowable-stress method, determine the maximum allowable eccentricity e if P= 26 kips. Use ay-36 ksi and E= 29 x 106 psi. (Round the final answer to three decimal places.) W4 x 13 The maximum allowable eccentricity e is[ Jinl

A steel compression member of 9-ft effective length supports an eccentric load as shown. Using the allowable-stress method, determine the maximum allowable eccentricity e if P= 26 kips. Use ay-36 ksi and E= 29 x 106 psi. (Round the final answer to three decimal places.) W4 x 13 The maximum allowable eccentricity e is[ Jinl

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

The lap joint is fastened with four ¾-inch diameter rivets. a. If the allowable shearing stress of the rivets is 14 ksi (kips per square inch), find the value of P (kips) that maybe applied to the joint. Assume that the load is equally distributed among the rivets. b. if the allowable bearing stress for the plates is 18 ksi, find the value of the load P (kips) that maybe applied to the joint. Assume that the load is equally distributed among the rivets. c. what is the maximum value of the load P (kips) that can be applied to the joint?
The compound axial member shown consists of a 20-mm diameter solid aluminum [E = 70 GPa] segment (1), a 24 mm-diameter solid aluminum segment (2), and a 16-mm diameter solid steel [E = 200 GPa] segment (3). Determine the displacement (mm) of member 3.
The bearings at A and D exert only y and z components of force on the shaft as shown in Fig 7. Determine the smallest diameter of the shaft using maximum shear stress hypothesis theory if the allowable stress is 100 MPa.
Solve for the indentation diameter if the BHN of the material is 70 and a 500 kg of load is applied? Use indenter diameter 10mm.
Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. What is the corresponding displacement at B in millimeter if P = 57 kN and Q = 397 kN?
A shaft is made of bronze at length AB and steel at length BC fixed at both ends. It is subjected to a concentrated torsional moment at the junction, The steel segment is 1 meter long. Assume that the Bronze has a diameter of 75mm, G = 35 GPa and τ = 60 MPa whilst the steel has a diameter of 50mm, G = 80 GPa, and τ = 80 MPa.a) What is the ratio of the length of steel to the length of bronze such that the allowable stress in each material is reached simultaneously.b) What is the reaction at the left end if the maximum torque is applied to the saft, in kN.
The rectangular bar has a width of w = 3.25 in. and a thickness of t = 2.00 in. The normal stress on plane AB of the rectangular block shown is 6 ksi (C) when the load P is applied. Assume a = 4 and b = 5. A) Determine the magnitude of load P. Express your answer in kips (rounded to the nearest hundredths.) B) Determine the shear stress on plane AB. Express your answer in ksi (rounded to the nearest tenths.)
The horizontal bar shown is pinned at O and is supported with a helical springat L1 from O. L1=2m, L2=1m. The spring has 30 turns and the diameter of the wire usedis 20mm. The diameter of the spring is 125mm. Modulus of Rigidity = 83GPa. Determine the deformation of the spring.
Two loads P act as shown, one parallel to (x0) and another parallel to z0; each load P equals 2.0 kN. The crankshaft dimensions are b1 = 90 mm, b2 = 120 mm, and b3 = 40 mm. The diameter of the upper shaft is d = 20 mm. Determine the normal stress at point B, which is located on the surface of the upper shaft at the y0 axis.
Two forces, each of magnitude P, are applied to the wrench. The diameter of the steel shaft AB is 30mm. Determine the largest allowable value of P if the shear stress in the shaft is not to exceed 120MPa and it's angle of twist is limited to 7deg. Use G=83GPa for steel. -Draw and label the diagram correctly, No diagram in the solution will be marked wrong. -Shortcut solution will be marked wrong.
The rigid beam BC is supported by rods (1) and (2). The cross-sectional area of rod (1) is 10 mm2. The cross-sectional area of rod (2) is 18 mm2. For a uniformly distributed load of w = 2.4 kN/m, determine the length a so that the normal stress is the same in each rod. Assume L = 5.25 m.
Link AB, of width b = 2 in. and thickness t = 1414 in., is used to support the end of a horizontal beam. The average normal stress in the link is -18 ksi and the average shearing stress in each of the two pins is 12 ksi. Determine the average bearing stress (ksi). **The way I had solved it was that I used the given normal stress to first find the P force needed to plus into my bearing stress formula. I still however need the diameter. The answer key says to use the given shear stress equation to find the diamter, how do you know to use that and not the shear stress formula again?