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 W24 X 55 8 ft 0 in 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 11 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 (2): Determine the governing load by LRFD and ASD methods for exterior floor system.
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 W24 X 55 8 ft 0 in 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 11 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 (2): Determine the governing load by LRFD and ASD methods for exterior floor system.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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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)
Hall
1 ft 0 in /
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
8 ft 0 in
80.
+
Solution:
floor =
D = Dselfweight of W24x55 + D building
L = L building floor = 8*80 = 640 lb/ft
I
W24 X 55
+
8 ft 0 in
8 ft 0 in
55+8*50= 455 lb/ft
[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. Wµ = 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, (W₁= 1570 lb/ft) is control to be used for design.
Homework (2): Determine the governing load by LRFD and ASD methods for
exterior floor system.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b1d03d5-ad01-4b6d-b55a-55b876b324cc%2F19f9d663-f95e-49d4-b3bb-646d2253409a%2Fj0mn6me_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)
Hall
1 ft 0 in /
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
8 ft 0 in
80.
+
Solution:
floor =
D = Dselfweight of W24x55 + D building
L = L building floor = 8*80 = 640 lb/ft
I
W24 X 55
+
8 ft 0 in
8 ft 0 in
55+8*50= 455 lb/ft
[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. Wµ = 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, (W₁= 1570 lb/ft) is control to be used for design.
Homework (2): Determine the governing load by LRFD and ASD methods for
exterior floor system.
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