Which apparatus is likely to carry a ladder? (There may be more than one answer.)
1 answer:
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Answer:
t=5.5 mm
Heat dissipation per unit length = 90.477 W/m
Explanation:
Given that
Diameter d = 5 mm ⇒r = 2.5 mm
Conductivity of insulated material K = 0.16 W/mK
Heat transfer coefficient = 20
When thickness reaches up to critical radius of insulation then heat dissipation will be maximum
We know that critical radius of insulation of wire is given as follow
Now by putting the values
So the thickness of insulation
t=8-2.5 mm
t=5.5 mm
As we know that heat transfer due to convection given as follows
Q = hAΔ T
Q=20 x 2 x π x 0.008 x (120-30)
Q = 90.477 W/m
So heat dissipation per unit length = 90.477 W/m
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.
