Answer:
provide 180 mm spacing
Explanation:
GIVEN DATA
rectangular beam: (b) = 300 mm, (d) = 575 mm
reinforced for flexure = 4Ф32 bars
WD = 30 kN /m, WL = 45 kN/m
Wu = 1.4 * 30 + 1.6 * 45 = 114 kN/m
i) concrete shear stress ( vc)
100 Ac / bd = (100 * u * 
* 32^2) / 300 * 575 = 1.865
from table 3.8
when: 100 Ac / bd = 1.865 then Vc = 0.778 N/mm^2
Ultimate shear force = (114 *5.5) / 2 = 313.5 kN
design shear stress = V / bd = (313.5 * 10^3) / (300 * 575) = 1.82 N/mm^2
v < 0.8
= 1.82 < 3.75
design link provided according to
Asv / sv = b(v-vc) / 0.87 fy = 300(1.82 - 0.778) / 0.87 (420)
ASv / Sv = 0.855
From table 3.13 :the value of Asv / sv can be calculated as
x = (-25) [ 0.625] + 200 = 184.375 mm
provide 180 mm spacing
Answer:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
Answer:
I am in 6th grade why am i in high school things.
Explanation:
Answer:
10kQ is to 36......... D2 _ D1 D4
Answer:
(a) See attachment
(b) The two planes are parallel because the intercepts for plane [220] are X = 0,5 and Y = 0,5 and for plane [110] are X = 1 and Y = 1. When the planes are drawn, they keep the same slope in a 2D plane.
(c) 
Explanation:
(a) To determine the intercepts for an specific set of Miller indices, the reciprocal intercepts are taken as follows:
For [110]
For [220]
The drawn of the planes is shown in the attachments.
(b) Considering the planes as two sets of 2D straight lines with no intersection to Z axis, then the slope for these two sets are:
For (1,1):
For (0.5, 0.5):
As shown above, the slopes are exactly equal, then, the two straight lines are considered parallel and for instance, the two planes are parallel also.
(c) To calculate the d-spacing between these two planes, the distance is calculated as follows:
The Miller indices are already given in the statement. Then, the distance is:
