Answer:
Shear resistance of a inclined stirrup is given by
Vᵇ = Asb(0.95*Fy)(cos α - sin α*cotβ) ((d-d')/sᵇ)
Explanation:
The shear resistance is checked in accordance with BS8110: Part 1, section 3.5.5. Shear resistance of a inclined stirrup is given by
Vᵇ = Asb(0.95*Fy)(cos α - sin α*cotβ) ((d-d')/sᵇ)
where,
Vᵇ=design shear resistance of inclined bars
Asb=cros-sectional area of the inclined bars
Fy=characteristics strength of the stirrups
α=angle between the inclined bars and the axis of the beam
β=angle between the compression strut of inclined bars and the axis of the beam
sᵇ=spacing of the inclined bars
d= effective depth and d'=effective depth minus cover to reinforcement
The shear stress is given by
v=V/bd
where V is the shear force due to ultimate loads. If v is shear stress
The Krebs cycle produces energy. Which starts with Glucose and ends in ATP. The only answer choice that begins with Glucose is A.
- Angle (θ) = 60°
- Force (F) = 20 N
- Distance (s) = 200 m
- Therefore, work done
- = Fs Cos θ
- = (20 × 200 × Cos 60°) J
- = (20 × 200 × 1/2) J
- = (20 × 100) J
- = 2000 J
<u>Answer</u><u>:</u>
<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
Explanation:
60 meters is he answer for this question
Eaither D or A bit I am leaning more towards D