Why 24kN is the maximum allowable value of P? PROBLEM 4Find the maximum allowable value of pand working Aresses aresteel:120=timber :2P x 10Concrete:500= 30 kN12 = P x 1032000= 24 kN16 = 5 P X 10800025. 6 KNshown in the figure.eTIMfor the column. ThePsteel:A = 500 mm²ew = 120 MPaTimber:A = 2000 mmOw = 12 MPaconcrete:2A 8000 mm ²Ow= 16MPa2P12p↓ P+5P↑ Spcross-sectional areas2Pal† P timber

Reference : Engineering Mechanics, Working stress. Given : Stress on each material (Steel, Timber…

PROBLEM 4 Find the maximum allowable value of p and working Aresses are steel: 120= timber : =1 Concrete: 2P x 10 500 = 30 kN 12 = P x 10 3 2000 = 24 kN 3 16 = 5 PX 10" 8000 25.6 KN for the column. The shown in the figure. be P steel: A = 500 mm tw = 120 MPa Timber: 2 A = 2000 mm Ow = 12 MPa concrete: A 8000 mm ² Ow= 16MPA £29 12p tp 45 P 15 5P cross-sectional areas 2P TP timber

PROBLEM 4 Find the maximum allowable value of p and working Aresses are steel: 120= timber : =1 Concrete: 2P x 10 500 = 30 kN 12 = P x 10 3 2000 = 24 kN 3 16 = 5 PX 10" 8000 25.6 KN for the column. The shown in the figure. be P steel: A = 500 mm tw = 120 MPa Timber: 2 A = 2000 mm Ow = 12 MPa concrete: A 8000 mm ² Ow= 16MPA £29 12p tp 45 P 15 5P cross-sectional areas 2P TP timber

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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