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4ar8 STD

The document is a project assignment from a College of Architecture course on steel and timber design. It provides two design problems involving calculating the load capacity of circular and square timber columns. For the first circular column, the document determines it is an intermediate column and calculates its effective stress, axial capacity, and verifies it is safe to support the given load. For the second square column, it determines it is a long column and calculates its effective stress and maximum supported load.

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JAVIN CALUYA
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576 views4 pages

4ar8 STD

The document is a project assignment from a College of Architecture course on steel and timber design. It provides two design problems involving calculating the load capacity of circular and square timber columns. For the first circular column, the document determines it is an intermediate column and calculates its effective stress, axial capacity, and verifies it is safe to support the given load. For the second square column, it determines it is a long column and calculates its effective stress and maximum supported load.

Uploaded by

JAVIN CALUYA
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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PONTIFICAL AND ROYAL

UNIVERSITY OF SANTO TOMAS


The Catholic University of the Philippines
COLLEGE OF ARCHITECTURE

CALUYA, Javin Alan F. March 21, 2023


4AR-8 ARC 1149: Steel and Timber Design

Project Assignment 2 / Quiz No. 2

1. A circular timber post 200 mm in diameter is to be used to support a load of 175 KN


Given that the allowable compressive stress is 9.6 MPa and E = 8240 Mpa, length is
2.0 m and buckling factor (ke) is 1.2, determine the following:
a. Whether the section is Short, Intermediate, or Long Column
b. The effective compressive stress
c. The axial capacity of the column.
d. Is the column safe to use?

Given: a. Determine column section


P = 175 KN 𝐿𝑒 𝐾𝑒 (𝐿)
𝑑
= 𝑑
Fc = 9.6 MPa = 9,600 KPa
1.2 (2 𝑚)
E = 8,240 MPa = 8,240,000 KPa = 0.2 𝑚
L = 2 meters
𝐿𝑒
Ke = 1.2 = 12
𝑑
d = 200 mm = 0.2 m
𝐸
K = 0.671
𝐹𝑐

8,240,000 𝐾𝑃𝑎
= 0.671
9,600 𝐾𝑃𝑎

K = 19.66

𝐿𝑒
Since 𝑑
= 12 is more than 11 but not

greater than K = 19.66, the column is an


Intermediate Column.
PONTIFICAL AND ROYAL
UNIVERSITY OF SANTO TOMAS
The Catholic University of the Philippines
COLLEGE OF ARCHITECTURE

b. Solving for the effective compressive stress for intermediate columns

𝐿𝑒 / 𝑑 4⎤
F’c = ⎡⎢1
⎣
−
1
3 ( 𝐾 )⎦
⎥
4
= F’c = ⎡⎢1
⎣
−
1
3 ( 12
)⎦
19.66 ⎥
⎤

F’c = 9,155.84 KPa

c. Column axial capacity


𝑃 𝑃
F’c =
𝐴
= π𝑑
2

𝑃
= F’c = 2
π(0,2 𝑚)
4

PCap = 287.64 KN

d. Is the column safe to use?


𝑃 𝑃
F’c =
𝐴
= π𝑑
2

175 𝐾𝑁
= F’c = 2
π(0,2 𝑚)
4

Fc = 5,570.42 KPa

9,155.84 𝐾𝑃𝑎
In conclusion, since the F’c = 287.64 𝐾𝑁
is greater than the actual compressive stress of the
5,570.42 𝐾𝑃𝑎
column which is Fc = 175 𝐾𝑁
, the column is SAFE.
PONTIFICAL AND ROYAL
UNIVERSITY OF SANTO TOMAS
The Catholic University of the Philippines
COLLEGE OF ARCHITECTURE

2. A timber post 300 mm x 300 mm is to be used to support a load P using Molave 80%
stress grade and a buckling factor (Ke) of 2.4. What is the load P? Height of the
column is 3.5 m.

Given:
Fc = 15.4 MPa
Ke = 2.4
L = 3.5 meters
E = 6.54 x 103 MPa = 6,450,000 KPa
d = 300 mm = 0.3 m

a. Determine column section


𝐿𝑒 𝐾𝑒 (𝐿)
𝑑
= 𝑑

2.4 (3.5 𝑚)
= 0.3 𝑚

𝐿𝑒
= 28
𝑑

𝐸
K = 0.671
𝐹𝑐

3
6.54 𝑥 10 𝑀𝑃𝑎
= 0.671
15.4 𝑀𝑃𝑎

K = 13.83

𝐿𝑒
Since 𝑑
= 28 is less than 50 and greater than K = 13.83, the column is a Long Column.
PONTIFICAL AND ROYAL
UNIVERSITY OF SANTO TOMAS
The Catholic University of the Philippines
COLLEGE OF ARCHITECTURE

b. Solving for the effective compressive stress for long columns

0.30 𝐸
F’c = 2
(𝐿𝑒 / 𝑑)

0.30 (6,450,000 𝐾𝑃𝑎)


= F’c = 2
(28)

F’c = 2,502.55 KPa

c. Computing for load PAct


𝑃 𝑃
F’c =
𝐴
= 2
𝑑

𝑃
= 2,502.55 KPa = 2
(0.3 𝑚)

PAct = 225.23 KN

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