Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
1 answer:
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
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Answer
given,
6 lanes divided highway 3 lanes in each direction
rolling terrain
lane width = 10'
shoulder on right = 5'
PHF = 0.9
shoulder on the left direction = 3'
peak hour volume = 3500 veh/hr
large truck = 7 %
tractor trailer = 3 %
speed = 55 mi/h
LOS is determined based on V p
10' lane weight ; f_{Lw}=6.6 mi/h
5' on right ; f_{Lc} = 0.4 mi/hr
3' on left ; no adjustment
3 lanes in each direction f n = 3 mi/h
= 0.877
= 1,555 veh/hr/lane
= (55 + 5) - 6.6 - 0.4 -3 -0
= 50 mi/h
level of service is D using speed flow curves and LOS for basic free moving of vehicle