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
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Why 24kN is the maximum allowable value of P?

Transcribed Image Text:PROBLEM 4
Find the maximum allowable value of p
and working Aresses are
steel:
120=
timber :
2P x 10
Concrete:
500
= 30 kN
12 = P x 10
3
2000
= 24 kN
16 = 5 P X 10
8000
25. 6 KN
shown in the figure.
e
TIM
for the column. The
P
steel:
A = 500 mm²
e
w = 120 MPa
Timber:
A = 2000 mm
Ow = 12 MPa
concrete:
2
A 8000 mm ²
Ow= 16MPa
2P
12p
↓ P
+5P
↑ Sp
cross-sectional areas
2P
al
† P timber
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