Answer:
Detailed solution is given in attached image
Find the toughness (or energy to cause fracture) for a metal that experiences both elastic and plastic deformation. Assume Equat
ion 6.5 for elastic deformation, that the modulus of elasticity is 172 GPa (25 ´ 106 psi), and that elastic deformation terminates at a strain of 0.01. For plastic deformation, assume that the relationship between stress and strain is described by Equation 6.19, in which the values for K and n are 6900 MPa (1 ´ 106 psi) and 0.30, respectively. Furthermore, plastic deformation occurs between strain values of 0.01 and 0.75, at which point fracture occurs.
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Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
<u>Determine the maximum speed for safe vehicle operation </u>
firstly calculate the stopping sight distance
m = R ( 1 - cos ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 +
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99