Structural Steel Design\Loads and Design Methods Example: The interior floor system shown has (W24x55) sections spaced (8ft) on center and is supporting a floor dead load of (50psf) and a live floor load of (80psf). Determine the governing load in (lb/ft) that each beam must support by using LRFD method. Area of slab supported per foot of length by W24 x 55 Area supported by one beam (shaded) W24 X 55 8 ft 0 in W24 X 55 4 ft 0 in 4 ft 0 in 8 ft 0 in I 1 ft 0 in 8 ft 0 in 1 W24 X 55 I W24 X 55 8 ft 0 in Solution: D = Dselfweight of W24x55 + D building floor = 55+8*50= 455 lb/ft L = L building floor = 8*80 = 640 lb/ft 8 ft 0 in [L=80psf < 100psf use factor in equations (3, 4 and 5) equal to (0.5)] [interior floor so not founded roof live, snow, or rain load] Lr or S or R = 0 1. W₁ = 1.4D = 1.4 *455 = 637 lb/ft 2. W₁ = 1.2D + 1.6L + 0.5(L, or S or R) = 1.2*637+1.6*640 = 1570 lb/ft 3. W₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*637+0.5*640 = 866 lb/ft 4. W₁ = 1.2D + 1.6W + 0.5 L + 0.5 (L, or S or R) = 1.2*637+0.5*640 = 866 lb/ft 5. W₁ = 1.2D ± 1.0E + 0.5 L + 0.2S = 1.2*637+0.5*640 = 866 lb/ft 6. W₁ = 0.9D (1.6W or 1.0E) = 0.9*455 = 409.5 lb/ft From above results, (W= 1570 lb/ft) is control to be used for design. Homework (1): Determine the governing load in (lb/ft) that each beam must support by using ASD method.
Structural Steel Design\Loads and Design Methods Example: The interior floor system shown has (W24x55) sections spaced (8ft) on center and is supporting a floor dead load of (50psf) and a live floor load of (80psf). Determine the governing load in (lb/ft) that each beam must support by using LRFD method. Area of slab supported per foot of length by W24 x 55 Area supported by one beam (shaded) W24 X 55 8 ft 0 in W24 X 55 4 ft 0 in 4 ft 0 in 8 ft 0 in I 1 ft 0 in 8 ft 0 in 1 W24 X 55 I W24 X 55 8 ft 0 in Solution: D = Dselfweight of W24x55 + D building floor = 55+8*50= 455 lb/ft L = L building floor = 8*80 = 640 lb/ft 8 ft 0 in [L=80psf < 100psf use factor in equations (3, 4 and 5) equal to (0.5)] [interior floor so not founded roof live, snow, or rain load] Lr or S or R = 0 1. W₁ = 1.4D = 1.4 *455 = 637 lb/ft 2. W₁ = 1.2D + 1.6L + 0.5(L, or S or R) = 1.2*637+1.6*640 = 1570 lb/ft 3. W₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*637+0.5*640 = 866 lb/ft 4. W₁ = 1.2D + 1.6W + 0.5 L + 0.5 (L, or S or R) = 1.2*637+0.5*640 = 866 lb/ft 5. W₁ = 1.2D ± 1.0E + 0.5 L + 0.2S = 1.2*637+0.5*640 = 866 lb/ft 6. W₁ = 0.9D (1.6W or 1.0E) = 0.9*455 = 409.5 lb/ft From above results, (W= 1570 lb/ft) is control to be used for design. Homework (1): Determine the governing load in (lb/ft) that each beam must support by using ASD method.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Steel Construction
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homework at the bottom of the page
![Structural Steel Design\Loads and Design Methods
Example: The interior floor system shown has (W24×55) sections spaced (8ft)
on center and is supporting a floor dead load of (50psf) and a live
floor load of (80psf). Determine the governing load in (lb/ft) that each
beam must support by using LRFD method.
Area of slab
supported per
foot of length
by W24 x 55 /
Area supported
by one beam
(shaded)
1 ft 0 in /
17
W24 X 55
W24 x 55 4 ft 0 in 4 ft 0 in W24 x 55
8 ft 0 in
8 ft 0 in
8coing
+
+
W24 X 55
+
8 ft 0 in
Solution:
D = Dselfweight of W24×55 + D building floor = 55+8*50 = 455 lb/ft
L = L building floor =
8*80 = 640 lb/ft
||
8 ft 0 in
[L=80psf < 100psf use factor in equations
(3, 4 and 5) equal to (0.5)]
[interior floor so not founded roof live, snow, or rain load]
Lr or S or R=0
1. W₁ = 1.4D = 1.4 *455 = 637 lb/ft
2. W₁ = 1.2D + 1.6L + 0.5(L, or S or R) = 1.2*637+1.6*640 = 1570 lb/ft
3. W₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*637+0.5*640 = 866 lb/ft
4. W₁ = 1.2D + 1.6W + 0.5 L + 0.5 (L₁ or S or R) = 1.2*637+0.5*640 = 866 lb/ft
5. W₁ = 1.2D + 1.0E + 0.5 L + 0.2S = 1.2*637+0.5*640 = 866 lb/ft
6. W₁ = 0.9D ±(1.6W or 1.0E) = 0.9*455 = 409.5 lb/ft
From above results, (W₁= 1570 lb/ft) is control to be used for design.
Homework (1): Determine the governing load in (lb/ft) that each beam must
support by using ASD method.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b1d03d5-ad01-4b6d-b55a-55b876b324cc%2F09e3c04f-13c9-4fbd-b47b-3e5bb5296d6e%2Fu0t095_processed.png&w=3840&q=75)
Transcribed Image Text:Structural Steel Design\Loads and Design Methods
Example: The interior floor system shown has (W24×55) sections spaced (8ft)
on center and is supporting a floor dead load of (50psf) and a live
floor load of (80psf). Determine the governing load in (lb/ft) that each
beam must support by using LRFD method.
Area of slab
supported per
foot of length
by W24 x 55 /
Area supported
by one beam
(shaded)
1 ft 0 in /
17
W24 X 55
W24 x 55 4 ft 0 in 4 ft 0 in W24 x 55
8 ft 0 in
8 ft 0 in
8coing
+
+
W24 X 55
+
8 ft 0 in
Solution:
D = Dselfweight of W24×55 + D building floor = 55+8*50 = 455 lb/ft
L = L building floor =
8*80 = 640 lb/ft
||
8 ft 0 in
[L=80psf < 100psf use factor in equations
(3, 4 and 5) equal to (0.5)]
[interior floor so not founded roof live, snow, or rain load]
Lr or S or R=0
1. W₁ = 1.4D = 1.4 *455 = 637 lb/ft
2. W₁ = 1.2D + 1.6L + 0.5(L, or S or R) = 1.2*637+1.6*640 = 1570 lb/ft
3. W₁ = 1.2D + 1.6(L, or S or R) + (0.5L or 0.8W) = 1.2*637+0.5*640 = 866 lb/ft
4. W₁ = 1.2D + 1.6W + 0.5 L + 0.5 (L₁ or S or R) = 1.2*637+0.5*640 = 866 lb/ft
5. W₁ = 1.2D + 1.0E + 0.5 L + 0.2S = 1.2*637+0.5*640 = 866 lb/ft
6. W₁ = 0.9D ±(1.6W or 1.0E) = 0.9*455 = 409.5 lb/ft
From above results, (W₁= 1570 lb/ft) is control to be used for design.
Homework (1): Determine the governing load in (lb/ft) that each beam must
support by using ASD method.
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