(a) Strain = compression / original length = 0.018/12 = 0.0015
(b) Axial compressive stress = force / cross sectional area = 200 kips / pi * (6/2)2 sq. inch = 200 /(3.141*9) kips = 7.073 kips/sq. inch.
(c) Axial compressive strength = force at the point of failure / cross sectional area
The required compressive strength is 4000 psi.
Calculated axial compressive stress = 7.07 kips/sq. inch. = 7.07 *1000 psi = 7073 psi.
Assuming that the concrete did not fail after applying 7073 psi of stress, this concrete mix can be approved.
A concrete test cylinder made from a known design mix is loaded with P-200 kips and compresses by...
MEE2001/2015 Q5. A thin walled steel cylinder with 10 mm wall thickness and internal diameter of 200 mm (Fig. Q5) is subjected to an internal pressure, p. The steel used for the cylinder has a yield stress of 240 MPa. +F 200 mm Fig. Q5 (a) If a tensile axial force of 1MN is applied to the end plate, show that the maximum internal pressure that can be applied before the cylinder yields according to the Tresca criterion is 16.2...