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

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

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Applications of Concrete Framing Systems

A concrete frame is a structure formed by the interconnection of beams and columns to form a structural network. In simple terms, it is the basic skeletal structure of a building or any construction. The concrete framing structure is built on a concrete foundation that forms the base of the entire structure. The concrete frame system is built to support floors, roofs, walls, beams, rafters, joists, etc.

Question

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

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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