A plastic cylinder is held snugly between a rigid plate and a foundation by two steel bolts (see figure). Steel L bolt Data for the assembly are as follows-length L = 600 mm, pitch of the bolt threads p = 1.0 mm, modulus of elasticity for steel E, = 195 GPa, modulus of elasticity for the plastic E, = 2.6 GPa, cross-sectional area of one bolt A, = 36.0 mm2, and cross-sectional area of the plastic cylinder A, = 960 mm2. Determine the compressive stress o, in the plastic when the nuts on the steel bolts are tightened by one complete turn. (Enter the magnitude in MPa.) 16905.96 X MPa

Q: 6.8 Problem: A bracket shown in figure 6.8 carries an eccentric load of 80 kN at an eccentricity o…

A:

Q: A steel [E = 200 GPa] rod with a circular cross section is 11 m long. Determine the minimum diameter…

A: Allowable stress=170 MPaForce=230 kN = 230000 NStress=ForceArea170=230000π4×d2d=41.5 mm

Q: The plates in the figure below are each 3 inches by 0.5 inches in cross section. If the hole in each…

A: Solution:- Given information: Cross section of plate =3×0.5 in Hole diameter, d=1316 in, and Load,…

Q: Problem 1) A rectangular ASTM A36 steel bar is 2in by 6in in cross section is subjected to an axial…

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Q: diameter of washer = 9/8 inch = stress in the nut 18 ksi g stress 800 psi

A: d=7/8 inch  dia of root thread = 0.731 inch din = 9/8 inch tensile stress = 18ksi bearing stress  =…

Q: An SAE3140 OQT 1300 steel bar with a rectangular cross section of  2 in ( base) by 3.5 in (height)…

A: From table Design Properties of Carbon and Alloy Steels (appendix 3 )   For SAE3140 OQT 1300…

Q: Calculate quantity of the Reinforcing for the beam shown in figure below. A 0.3 m 3.2 m 2016 08 @…

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Q: What is the approximate value of the tensile strength governed by yielding of the cross- section of…

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Q: Two 20 mm thick slabs are joined by a 10 mm thick slab as shown in the figure. The bolts used as…

A: The thickness of thinner plate=10 mm The thickness of the longer plate=20 mm Diameter=5/8 inch

Material Properties

Construction in civil engineering demands high strength, durability, and longevity. These targets can be achieved by better design and analysis, along with the correct choice of materials. Different materials have different properties and hence, they respond and performs differently under the various condition of external loadings. It is therefore a critical task of an engineer to select the right material based on the property that the material is supposed to function in real-time. When speaking about the property, there are different categories of material properties like chemical property, physical property, mechanical property, electrical property, magnetic property, thermal property, etc. These material properties form the basis of material selection in various applications.