Why did it became 4F/A from F/A, in (a) DESIGN PROBLEMS36.Solution:s = axial stresss = shear stress(a)A hook is attached to a plate shown and supports a static load of 12,000 lb. Thematerial is to be AISI C1020, as rolled. (a) Set up strength equations fordimensions d, D, h, and t. Assume that the bending in the plate is negligible.(b) Determine the minimum permissible value of these dimensions. In estimatingthe strength of the nut, let D₁ = 1.2d. (c) Choose standard fractional dimensionswhich you think would be satisfactory.S=F・nd²4FπdRound NutE==CHAHEELCentradace²Problems 36-38.

Answered: S= 1 4 F nd² = 4F πα?

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Mechanical Engineering

S= 1 4 F nd² = 4F πα?

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.6.5P: An aluminum bar has length L = 6 ft and diameter d = 1.375 in. The stress-strain curse for the...

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Question

Why did it became 4F/A from F/A, in (a)

DESIGN PROBLEMS
36.
Solution:
s = axial stress
s = shear stress
(a)
A hook is attached to a plate shown and supports a static load of 12,000 lb. The
material is to be AISI C1020, as rolled. (a) Set up strength equations for
dimensions d, D, h, and t. Assume that the bending in the plate is negligible.
(b) Determine the minimum permissible value of these dimensions. In estimating
the strength of the nut, let D₁ = 1.2d. (c) Choose standard fractional dimensions
which you think would be satisfactory.
S=
F
・nd²
4F
πd
Round Nut
E
==
CHA
HEEL
Centrad
ace²
Problems 36-38.

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