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Timber Timber and Steel Design and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design of Beams for Shear Deflection Design for Bearing Floor System Design Timber Timber 3: 3: Timber Beam Timber Beam Mongkol JIRAVACHARADET S U R A N A R E E INSTITUTE OF ENGINEERING UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

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Page 1: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

TimberTimber and Steel Designand Steel Design

� Flexure-Laterally Supported Beams

� Flexure-Laterally Unsupported Beams

� Design of Beams for Shear

� Deflection

� Design for Bearing

� Floor System Design

Timber Timber 3:3: Timber BeamTimber Beam

Mongkol JIRAVACHARADET

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

Page 2: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Wood Floor System

Lateral Support of Beams

Page 3: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Flexural Resistancebb Ff ≤

bF

Mbd

62 =

Shear Resistancevv F

bd

Vf <=

2

3

Bearing Resistance⊥⊥ <= c

b

c FA

Rf

Page 4: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Decking Patterns

EI

wL

wLM

384

5 max

8 max

4

2

=∆

=

EI

wL

wLM

185 max

8 max

4

2

=∆

=

EI

wL

wLM

109 max

8 max

4

2

=∆

=

Page 5: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

EI

wL

wLM

145 max

10 max

4

2

=∆

=

EI

wL

wLM

105 max

10 max

4

2

=∆

=

Page 6: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Laterally Unsupported Beam

Page 7: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Effective Length

�������

�����������

���� ��

������ ����� ����������������� le����������������������

��������������������

1.3lu + 3d

1.84lu

�������������������

��������!""���#$

1.63lu + 3d

1.84lu ����"�! lu/d > 14.31.63lu + 3d ����"�! lu/d < 14.3

�������������������

�������������������� &�"'

��������!""���#$

0.90lu + 3d

1.44lu + 3d

1.84lu ����"�! lu/d > 14.31.63lu + 3d ����"�! lu/d < 14.3

Page 8: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Short, Intermediate, and Long Beams

Slenderness factor: 2/bdlC es =

bb FF =′Short Beams: Cs < 10

Intermediate Beams: 10 < Cs < Ck

bk FEC /811.0=b

k

sbb F

C

CFF ≤

−=′

4

3

11

Long Beams: Cs > Ck

bsb FCEF ≤=′ 2/438.0

Page 9: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

38.15)5.554.2/()5.1754.2)(039,1(/ 22 =××== bdlC es

16.25120/464,115811.0/811.0 === bk FEC

�� � ���� 10 < Cs < Ck �����������������

����������� 21.1 ���(�)��*���������+�� 6 x 18 �&�� ���������� 12 ���" ������"" �"�!)��+)����������'������������������� ���"��� !�1����"� ��������!""����������� �����"�!()��������#?

������ �����"�� �.1 (�)��*� Fb = 120 ��./7�.2 ��' E = 115,464 ��./.7�.2

����������� = 1,070(6)(8)(2.54)2/1002 = 33 ��./���"

lu = 6(100) = 600 7�.

le = 1.63(600) + 3(8)(2.54) = 1,039 7�. (��"���� 21.1)

Page 10: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

K ksc 120 ksc 4.11416.25

38.15

3

11120

4

O<=

−=′bF

w = 292.4 ��./#���

����������� �#�) M = F’b S

w(12)2/8 = 114.4(4,600)/100

Page 11: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Critical Sections for Shear

d

d

Support with Bearing

Critical section

Face of support

d’

Support without Bearing

Critical section

Face of support

Page 12: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

�� � �������AB�" �"�!����"1� ��)����&�C��' �B���"'�' d

3.0 t 3.0 t

0.44 m 1.5 m 3.06 m

5353.5 kg 2209.5 kg

312.6 kg/m

����������� 21.2 ���(�)��+�� 10 x 18 �&�� ������� 5 ���" �������"�!�"��!����������'���� 10 7�. "�!��������!""������� 200 ��./���" !����������������� �(����A�������1�' 1.05) ���������"� �) +�� 3.0 ���� ��) "'�'���� 1.5 ���" ���"��� !����������"A"�!�"��D �()�" (��?

������ �����"�� �.1 ������"��D �+����������� �#�)+ �(�)�� Fv = 12 ��./7�.2

����������� = (9.5)(17.5)(2.54)2(1,050)/1002 = 112.6 ��./���"

��������������������� = 200 + 112.6 = 312.6 ��./���"

Page 13: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

fv = 1.5V/A = 1.5(5,216)/(9.5)(17.5)(2.54)2

= 7.29 ��./7�.2 < [Fv = 12 ��./7�.2] OK

�"��D �����)����&�C� V = 5353.5 - 0.44(312.6) = 5,216 �&���"��

������"��D �����&+E��:

5353.5 kg

Shear Diagram

d = 0.44 m

Page 14: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

DLLLTotal ∆+∆=∆ 5.1Deflections

�������� 21.2 ��"� �������� �#�)+ ����(�)

���$%����� ������

�����&�

������ ����������

'�������� �����&�

��������� ���" ������""� L/180 L/120 ���"1��&�����B��� D�!+)������ L/360 L/240 ���"1��&�(����B��� D�!+)������ L/240 L/180

���"�!1 �� L/360 L/240

Page 15: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

�������� 21.3 "'�'� ������������

���$��� ���'������&�����$�$

���'������&�����#)*��

�����������!������������ 5wL4/384EI 1.2wL

2/8AG

�������������������������� PL3/48EI 1.2PL/4AG

���������������������� �� PL3/3EI 1.2PL/AG

�����������!����� �� wL4/8EI 1.2wL

2/2AG

Page 16: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Max. Beam Bending Moments, Shears,

and Deflections

0.125wL2 0.5wL 5wL4/384EI

Max. M Max. V Max. ∆

0.125wL2 0.625wL wL4/185EI

0.1wL2 0.6wL wL4/145EI

0.107wL2 0.607wL wL4/154EI

Page 17: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Design for Bearing

⊥≤ cb FAR /

Ab = Bearing area = b � lb

bFRl cb ⊥= /

Page 18: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

������ �����"���� �.1 Fg = 90 ��./7�.2 ��' Fc = 30 ��./7�.2

Bearing at an Angle to Grain

θθ 22 cossin

+

×=

cg

cg

nFF

FFF

ksc 4.4150cos3050sin90

)30)(90(22

=°+°

=nF

Page 19: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

Floor System Design

30 cm

25 m

5.4 m6 m

N

����������� 21.4 �� ��!!"'!!1 ������"�!�����&��)�#�"B��� 21-10 ��������!""����" = 500 ��./�"�. ��'��������!""������� = 50 ��./�"�. !����������(�) ���� &I�� �������� 30 7�.

������ ����� �+ �"'!!���/��()���

1. ���#��&J N-S ������� 5.4 ���" ��#���� E-W

2. ���#��&J N-S ������������������ ��#���� E-W

3. ���#��&J E-W ��������" �"�! ��#���� N-S

��"#�)������*�"�����!��(�)�* ������ ������ ���E��

Page 20: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

������' ,�-.*��

�� �(�)�'�!���� 2 �&����������(�)�"'��

�����O����"�!�A!1 ����)�� 1 ���"

�������������+ �(�)�'�!� 720 ��./�!�.

�����&���(���) ���"������� �������E��� ������ �����E��

�����"���� �.3 ����"�!��������� Mmax = 0.107wL2

w = 500 + 50 + 720(1.5)(2.54)/100 = 577 ��./�"�.

������"����� �#�) Fb = 100 ��./7�.2

���������� �#�) M = 100((100)(1.5 � 2.54)2/6)/100 = 242 ��.-�./�.

m 98.1)107.0577/(242 =×=L�������������� �#�)

Page 21: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

)461)(556,112(154

)100()577(5.1

240

100 34LL=

�������������� �#�) L = 1.57 ���" (�� ���)

���+�������"� ����� L/240 1&��"O��������������� �#�)

I = (100)(1.5 � 2.54)3/12 = 461 7�.4/���"

�����"���� �.3 ����"�!��������� ∆max = wL4/154EI

1&��"O���+ ������ ! ∆max = 1.5wL4/154EI

Page 22: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

fv = 1.5(419)/(1.5)(2.54)(100)

= 1.65 ��./7�.2 < [Fv = 10 ��./7�.2] OK

���&�� '��#)*��1�,�-���$�� A)�����#�)�������)�� 2 �&��

"'�'�� ����������� = 1.57 S (1.5)(2.54)/100 = 1.53 ���"

1.57 m

2”

�"��D ��&�C� V = 577(1.53/2 S (1.5)(2.54)/100)

= 419 ��./������)�� 1 ���"

Page 23: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

����������� ������� = 60 ��. c/c

�"'��O���������� = 10 ��./���"

w = 577(0.60) + 10 = 356 ��./���"

M = 356(2.76)2/8 = 339 ��.-���"

������' �� �������+ � ���"���������������! 24.4 ���"

� �#�)"'�'������� 3 ���" 8 ���� = 24 ���"

�����&�����)�� 10 �&�� �� �����"'������� = 3 S 9.5(2.54)/100 = 2.76 ���"

������"����� �#�) Fb = 100 ��./7�.2

������) ���" S = 339(100)/100 = 339 7�.3

�-���$���#�2�,�,$-: 2 x 10 ���� (S = 369 9�.3, ��� 6.6 ��./�.)

Page 24: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

��)����'#�)����������������� ���(�)���+����*�����������

#�*��1�-,�-��' � ��������� = 60 9�. c/c

����������� ������� = 100 ��. c/c

�"'��O���������� = 15 ��./���"

w = 577(1.0) + 15 = 592 ��./���"

M = 592(2.76)2/8 = 564 ��.-���"

������"����� �#�) Fb = 100 ��./7�.2

������) ���" S = 564(100)/100 = 564 7�.3

��)���������(�(): 3 x 10 �&�� (S = 616 7�.3, ���� 11 ��./�.)

Page 25: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

���&�� ���'������%����:

���&�� �������' ����%����: Fc = 22 ��./7�.2

w = 577(0.6) + 6.6 = 353 ��./���"

R = 353(3/2) = 530 ��.

lb = 530/(22)(1.5)(2.54) = 6.32 7�.

cm 45.0)913,7)(556,112(384

)100()76.2)(353)(5(5.1384/)5(5.1

344 ===∆ EIwL

= L/614 OK

Page 26: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

���&�� '��#)*��1���:

�"��D ��&�C� V = 353(2.76/2 S (11.5)(2.54)/100)

= 384 ��.

fv = 1.5(384)/(1.5)(11.5)(2.54)2

= 5.18 ��./7�.2 < [Fv = 10 ��./7�.2] OK

������' ��� "'�'����"'��������� &I�� = 5.4 ���"

������������� = 3(353)/0.60 = 1,765 ��./���"

����������� = 100 ��./���"

�������������� = 1,865 ��./���"

M = 1,865(5.4)2/8 = 6,798 ��.-���"

Page 27: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

� �#�)(�)�'�!� ������"����� �#�) Fb = 100 ��./7�.2

������) ���" S = 6,798(100)/100 = 6,798 7�.3

�� ���)���: 10 x 18 �&�� (S = 7,946 7�.3, ���� 77 ��./���")

���&�� �������' ����%�����: Fc = 22 ��./7�.2

w = 1,765 + 77 = 1,842 ��./���"

R = 1,842(6/2) = 5,526 ��.

lb = 5,526/(22)(9.5)(2.54) = 10.4 7�.

Page 28: Timber and Steel Designeng.sut.ac.th/ce/oldce/CourseOnline/430432/L22 Timber3.pdfTimber and Steel Design Flexure-Laterally Supported Beams Flexure-Laterally Unsupported Beams Design

���&�� ���'������%�����:

cm 5.1)600,176)(556,112(384

)100()4.5)(842,1)(5(5.1384/)5(5.1

344 ===∆ EIwL

���&�� '��#)*��1����:

�"��D ��&�C� V = 1,842(5.4/2 S (17.5)(2.54)/100)

= 4,155 ��.

fv = 1.5(4,155)/(9.5)(17.5)(2.54)2

= 5.81 ��./7�.2 < [Fv = 10 ��./7�.2] OK

= L/351 OK