Problem 2. The length of a side of the square block is 4 in. Under the application of the load V, the top edge of the block disp
laces 1 16 in. (dashed lines show displacement). Determine the shear strain at corner A and the shear strain at angle COD.
1 answer:
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Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + ..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 +
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - ) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - )
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = ..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 =
solve it we get
e = 2%