Thanks for the help!
1 answer:
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Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar =
ar =
ar = 22.74 m/s²
so magnitude of total acceleration is
A =
A =
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²
Answer: ASD = 306 kips-ft
LRSD = 1387.5 k-ft
Explanation:
To begin, we will take a step by step process to solving this problem.
Attached below is a picture to guide us to solving this.
To begin, we have that to reaction of the support
ΣMд = 0
where;
RB * 35 - (8+18)15 - (4+9)20 = 0
RB = 18.57k
also Ey = 0;
RA + RB = 18 + 8 + 9 +4 = 20.43 k
taking the maximum moment at mid point;
Mc = RA * 35/2 - (8 +18) (35/2 -15)
Mc = 292.525
therefore, MD = RA * 15 = 20.43 * 15 = 306.45 k.ft
MD = 306.45 k.ft
ME = 279 k.ft i.e 18.57 * 15
considering the unsupported length; 35 - (15*2 = 5ft
now we have that;
Lb = Lp = 5ft
where Lp = 1.76 ry(√e/fy)
Lp = 1.76 ry √29000/50 ......
ry = 1.4 inch
so we have that Mr = Mp for Lb = Lp where
Mp = 2 Fy ≤ 1.5 sx Fy
Recall from the expression,
RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6
RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20
RA = 30.17 k
the maximum moment at D = 30.17 * 15 = 452.55 k.ft
Zrequired = MD/Fy = 452.55 * 12 / 50 = 108.61 inch³
so we have Sx = 452.55 * 12 / 1.5 * 50 = 72.4 inch³
also r = 1.41 in
Taking LRFD solution:
where the design strength ∅Mn = 0.9 * Zx * Fy
given r = 2.97
Zx = 370 and Sx = 81.5, we have
∅Mn = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 k-ft
this tells us it is safe.
ASD solution:
for Lb = Lp, and where Mn = Mp = Fcr Sx
we already have value for Sx as 81.5 so
Fcr = ZxFy/Sx
Fcr = 370 * 50 / 81.5 = 227 ksi
considering the strength;
Strength = Mn / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-ft
This justifies that it is safe because is less than 306
cheers i hope this helps.
