MAT. SCI. & ENG: AN INTO. WILEYPLUS
MAT. SCI. & ENG: AN INTO. WILEYPLUS
10th Edition
ISBN: 9781119472001
Author: Callister
Publisher: WILEY
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Chapter 8, Problem 6QAP
To determine

The fracture occurs or not at given stress and crack length and reason in support of the answer.

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1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²) simply supported slab shown in Fig. Solution: 2 m 2 m هنا الاسناد بسيط، لذلك سيتشكل خط خضوع واحد بالمنتصف ( البلاطة متناظرة) = We [5.0x (2x1.5) 0 = 8/2 :. W;= [m × 8/2 × 1.5] <2 = [1.5m 6] :: We = Wi 15 6 = 1.5 m 6 m = 10 kN.m 8/2] -8=1.0 1.5 m E E L 8/2 δ 28 0 = L/2 L
An AISI 1018 steel ball with 1.100-in diameter is used as a roller between a flat plate made from 2024 T3 aluminum and a flat table surface made from ASTM No. 30 gray cast iron. Determine the maximum amount of weight that can be stacked on the aluminum plate without exceeding a maximum shear stress of 19.00 kpsi in any of the three pieces. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pmax 0.4 90 0.6 στ Tmax 0.2 0.5a a 1.5a 2a 2.5a За Distance from contact surface The maximum amount of weight that can be stacked on the aluminum plate is lbf.
A carbon steel ball with 27.00-mm diameter is pressed together with an aluminum ball with a 36.00-mm diameter by a force of 11.00 N. Determine the maximum shear stress and the depth at which it will occur for the aluminum ball. Assume the figure given below, which is based on a typical Poisson's ratio of 0.3, is applicable to estimate the depth at which the maximum shear stress occurs for these materials. 1.0 0.8 Ratio of stress to Pma 9 0.6 στ 24 0.4 Tmax 0.2 0 0.5a a 1.5a Z 2a 2.5a За Distance from contact surface The maximum shear stress is determined to be MPa. The depth in the aluminum ball at which the maximum shear stress will occur is determined to be [ mm.
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