Problem 1: Load Distribution The diagram is to support a 100 mm thick concrete slab. 5.8 kPa of dead load and 2.1 kPa of live load will be overlaying on the slab. The following is the cross-section of the beams: AB, GH = 250mm X 450mm AG, BH = 250mm X 450mm CD, EF = 250mm X 550mm a) Self weight = (0.25x0.55x5.4) x 15 = 11.1375 kN-A Tributary area = 5.4x1.25+5.4x1.5 =14.85m^2 Self load = (14.85x0.1) x 15 B = 22.275KN Dead load =14.85x5.8 =86.13KkN Total dead load =A+B+C - 110 54 5.4m - - 3m 2.5m 3m Determine the following if the unit weight of concrete is 15 kN/m3 a. b. C. b) Live load Dead loads acting on beam EF. Live load acting on beam CD. Dead load acting on girder AG. = 14.85x2.1 = 32. 185 KN Total live load = 32.185 c) Self weight = (0.25x0.45x8.5) x 15 = 34.425 kN-A Tributary area = 2.7x8.5 = 22.95m^2 Self load = (22.95x0.1) x 15 = 34.425kN-B Dead load = 22.95x5.8 = 133.11 KN-C Total dead load = A + B + C =161.88 kN =I ald

Structural Analysis
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Chapter2: Loads On Structures
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Qno8 solve like sample check I don't have problem for 8 . Solve problem 8 information with figure.
Problem 8: Approximate Analysis for Vertical Loads
Transcribed Image Text:Problem 8: Approximate Analysis for Vertical Loads
Problem 1: Load Distribution
The diagram is to support a 100 mm thick concrete slab. 5.8 kPa of dead load and 2.1 kPa of live
load will be overlaying on the slab. The following is the cross-section of the beams:
AB, GH = 250mm X 450mm
AG, BH = 250mm X 450mm
CD, EF = 250mm X 550mm
a)
Self weight
= (0.25x0.55x5.4) x 15
= 11.1375 kN-A
Tributary area
= 5.4x1.25+5.4x1.5
=14.85m^2
Self load
= (14.85x0.1) x 15
B
= 22.275KN
Dead load
=14.85x5.8
=86.13KN
5.4m
-
= 119.54kN
-
Total dead load
= A + B + C
3m
2.5m
3m
Determine the following if the unit weight of concrete
is 15 kN/m3
a.
b.
C.
b)
Live load
Dead loads acting on beam EF.
Live load acting on beam CD.
Dead load acting on girder AG.
= 14.85x2.1
= 32. 185 kN
Total live load = 32.185
c)
Self weight
= (0.25x0.45x8.5) x 15
= 34.425 kN-A
Tributary area
= 2.7x8.5
= 22.95m^2
Self load
= (22.95x0.1) x 15
= 34.425kN
B
-
Dead load
= 22.95x5.8
= 133.11 KN-C
Total dead load
= A + B + C
=161.88 kN
=I
Transcribed Image Text:Problem 1: Load Distribution The diagram is to support a 100 mm thick concrete slab. 5.8 kPa of dead load and 2.1 kPa of live load will be overlaying on the slab. The following is the cross-section of the beams: AB, GH = 250mm X 450mm AG, BH = 250mm X 450mm CD, EF = 250mm X 550mm a) Self weight = (0.25x0.55x5.4) x 15 = 11.1375 kN-A Tributary area = 5.4x1.25+5.4x1.5 =14.85m^2 Self load = (14.85x0.1) x 15 B = 22.275KN Dead load =14.85x5.8 =86.13KN 5.4m - = 119.54kN - Total dead load = A + B + C 3m 2.5m 3m Determine the following if the unit weight of concrete is 15 kN/m3 a. b. C. b) Live load Dead loads acting on beam EF. Live load acting on beam CD. Dead load acting on girder AG. = 14.85x2.1 = 32. 185 kN Total live load = 32.185 c) Self weight = (0.25x0.45x8.5) x 15 = 34.425 kN-A Tributary area = 2.7x8.5 = 22.95m^2 Self load = (22.95x0.1) x 15 = 34.425kN B - Dead load = 22.95x5.8 = 133.11 KN-C Total dead load = A + B + C =161.88 kN =I
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