The uniform service dead load is 1.0 k/ft and a live load of 1.5 k/pt on a simpl
ID: 1709861 • Letter: T
Question
The uniform service dead load is 1.0 k/ft and a live load of 1.5 k/pt on a simple span of 30 ft. Assume that the section is fully composite and the concrete has a density of 150 pcf with a 28 day compressive strength of 3.00 Ksi. Use ASTM A steel with F_y = 5 Ksi, For the composite section shown, determine: a. The transformed moment of Inertia. b. The design strength of the composite section. c. If the section is adequate to carry the given loads. Design a base plate supporting a W 8 times 31 column, supported on a -0 times 1' - 6" footing. P_U = 150 Kips, f'_c = 3 ksi, and F_y = 36 ksi. Determine if the dead load deflection in problem 1 is within the allowable limit. Delta_allow = L/240. Check steel members only.Explanation / Answer
Uniform service load = 1.0k/ft
Live load = 1.3k/ft
Span length = 30 ft
Composite section
Concrete density = 150 pcf
28 day compressive strength = 3 ksi
ASTM steel A36 , Fy = 36 Ksi
bE = 90”, ts = 2.5 “", W16x26 steel beam, 3/8"x6(1/2)" steel cover plate, normal weight concrete with a 28-day strength (f'c) = 3 ksi
1.Transform the concrete into equivalent steel.
1.1 find Ec so that you can compute n. According to ACI 318, Ec
Ec = 33 w1.5 sqrt(f'c) = 33 (150 pcf)1.5 sqrt(3,000 psi)
Ec = 3.320 x106 psi
Now we compute n:
n = 29 / 3.320 ~ 8.734
To transform the section into all steel, we need to divide the concrete dimension parallel to the axis (i.e. bE) by n and draw the new section.
bE / n = 90" / 8.734 = 10.30"
W 16 X 26 features:
d= 15.59 in.
Ix= 301 in4
Ag= 7.68 in2
Find Neutral Axis
Part
Area
y top
YA
in2
in
in3
Concrete(10.30 X 2.5)in2
25.75
1.25
32.1875
Beam
7.68
13.295
102.1056
Cover PL
2.43
21.84
53.0712
Whole Section
35.86
187.3643
NA
5.225
From the top of the slab
Use Parallel axis theorem for ENA location
Part
Area
y(NA)
(Y^2)A
Io
ITR(in^4)
Concrete(10.30 X 2.5)in2
25.75
-3.975
406.8421
13.411
420.2531
Beam
7.68
8.070
500.1738
301
801.1738
Cover PL
2.43
16.615
670.8309
0.2285
671.0594
Whole Section
1892.486
Total weight = DL + LL = (1.3 +1.0)k/ft = 2.3 k/ft
LFRD
ASD
Wu = (1.2 x 1.0 + 1.6 X 1.3) k/ft= 3.28 k/ft
Mu = (3.28 k/ft (30ft)^2)/8 = 369 k-ft
Wa =( 1+1.3)k/ft = 2.3 k/ft
Ma = (2.3 k/ft (30ft^2))/8 = 258.75 k-ft
2.Is the section safe to carry the load
Determine beff
The effective width of the concrete slab is the sum of the effective widths for each side of the beam centerline, which shall not exceed:
(30/8)ft. 2 = 7.5 ft.
Calculate the moment arm for the concrete force measured from the top of the steel shape, Y2. Assume a = 1.0 in. (Some assumption must be made to start the design process. An assumption of 1.0 in. has proven to be a reasonable starting point in many design problems.)
Y = 10.545 in-1 in = 9.545
LFRD
ASD
Mn = 1.12 X 369 > 369
Mn/ = (258.75/0.95) >258.75
Hence Safe
Find Neutral Axis
Part
Area
y top
YA
in2
in
in3
Concrete(10.30 X 2.5)in2
25.75
1.25
32.1875
Beam
7.68
13.295
102.1056
Cover PL
2.43
21.84
53.0712
Whole Section
35.86
187.3643
NA
5.225
From the top of the slab
Use Parallel axis theorem for ENA location
Part
Area
y(NA)
(Y^2)A
Io
ITR(in^4)
Concrete(10.30 X 2.5)in2
25.75
-3.975
406.8421
13.411
420.2531
Beam
7.68
8.070
500.1738
301
801.1738
Cover PL
2.43
16.615
670.8309
0.2285
671.0594
Whole Section
1892.486
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