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1 Strength (ultimate strength) design method Flexural Analysis and Design of Beams:

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Page 1: 3. Strength design method

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Strength (ultimate strength) design method

Flexural Analysis and Design of Beams:

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U�����������Safety factors U

U������������ ������������������������������U :

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Ex.1: A 2.4m span cantilever beam has a rectangular section of b =200mm and d = 390mm with 3 bars of 22mm diameter, carries a uniform dead load including it’s own weight of 12kN/m and a uniform distributed live load of 10.5kN/m. Check the adequacy of the section, using f’c of 28Pa and fy of 280MPa?

�� = 1.2 � 12 + 1.6 � 10.5 = 31.2��/��� = �� 22 = 31.2 � (2.4)22 = 89.86 ��. ��� = ���� = 3 � � � 2224 � 200 � 390 = 0.0146��� = ����4��� � 1.4�� , �� � = �284�280 � 1.4280 ,

��� = 0.0047 � 0.005 ��� ��� = 0.005��� = 0.75� = 0.75 � 0.85 � 0.85 � 28280 � 600600 + 280 = 0.0362 ��� ��� � �� � ��� the beam is under reinforced o.k.

� = �� � ��0.85 � �� � � � = 1140 � 2800.85 � 28 � 200 = 67��

�� = � �� � �� !� " �2#

= 0.9 � 1140 � 280 � !390 " 672 # = 102.4 ��. � > 89.86��. � ����$ ��%�

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Ex.2: Determine the allowable moment and the position of neutral axis of a rectangular section which has b = 200mm, d =300mm use f’c= 21MPa and fy= 280MPa, when:

a) Reinforced with 3Ø 20 mmb) Reinforced with 3Ø 28mm

Solution:a) �� = ���� = 3���2024�200�300 = 0.0157��� = ����4��� � 1.4�� , ��� = �214�280 � 1.4280 ,��� = 0.00409 � 0.005 ��� ��� = 0.005��� = 0.75�= 0.75 � 0.85 � 0.85 � 21280 � 600600 + 280= 0.0277 ��� ��� � �� � ��� then beam is under reinforced o.k.�� = ���2�� &1 " 0.59 ���� �' =�� = 0.9 � 0.0157 � 200 � 300 � 300 � 280 !1 " 0.59 � 0.0157 28021 #106 = 62.4��. �� = �� � ��0.85 � �� � � � = 942 � 2800.85 � 21 � 200 = 74��

� = �* = 740.85 = 87.06��

b) �� = ���� = 3���2824�200�300 = 0.0308��� = ����4��� � 1.4�� , ��� = �214�280 � 1.4280 ,��� = 0.00409 � 0.005 ��� ��� = 0.005��� = 0.75� = 0.75 � 0.85 � 0.85 � 21280 � 600600 + 280 = 0.0277 ��� ��� � �� � ��� then beam is over reinforced then use �������� = 0.0277 � 200 � 300 = 1662.6 ��2�� = ���2�� &1 " 0.59 ���� �'�� = 0.9 � 0.0277 � 200 � 300 � 300 � 280 !1 " 0.59 � 0.0277 28021 #106 = 98.27��. �� = �� � ��0.85 � �� � � � = 1662.6 � 2800.85 � 21 � 200 = 130.4�� � = �* = 130.40.85 = 154��

For this case fs are less than fy.

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Ex.3: Design the simply supported rectangular beam with span of 4m support service dead load of 10kN/m and service live load of 30kN/m f’c=30MPa and fy = 400MPa,

�������������� !"����#$���#�

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Ex.4: simply supported beam with 6m span, support a service dead load of 10 N/m and point live load at mid span 50kN, Design the beam using f’c = 30 MPa and fy= 350 MPa try =0.6 of ��� ?

���� = 16 (0.4 + ��700)���� = 600016 !0.4 + 350700# = 337.5�� , 340Assume b= 250mm then Wd= 0.25*0.34*24 = 2.04 kN/m

Total load= 10+4=14

�� = 1.2�14�628 + 1.6�50�64 = 195.6�. ���� = 0.75� = 0.75 � 0.85 � 0.85 � 30350 � 600600 + 350 = 0.02933 = 0.6 � 0.02933 = 0.0176��2 = ������(1 " 0.59 �����) = 195.6 � 106

0.9 � 300 � 0.0176 � 350(1 " 0.59 � 0.0176 35030 )��2 = 40144793.16 ��3 � = -40144793 .16�

�(�� ) �(��) ��200 448 2.24250 400 1.6300 365.8 1.219

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Use b = 200mm and d= 448 or 450 mm then h= 450 +70= 520mm

Wd= 24*0.20 *0.52= 2.6 kN/m < 4 kN/m o.k.

As= 0.0176*200*450 = 1584 mm

Ex.5: Find the minimum dimension of the cross section for the beam shown with b=300mm , Use f’c= 25MPa and fy= 400 MPa, then find area of steel for the whole beam?

�� = 1.2 � 30 + 1.6 � 18 = 64.8 ��/���� = ����4��� � 1.4�� ,

��� = �254�400 � 1.4400 ,

��� = 0.00315 � 0.0035 ��� ��� = 0.0035��� = 0.75� �:; ���. ��� �:� ��� ���= 0.75 � 0.85 � 0.85 � 25400 � 600600 + 400 = 0.0205

� = < ��� � � � � �� � (1 " 0.59 � ���� �)

� = < 302.19 � 1060.9 � 0.0205 � 300 � 400 � (1 " 0.59 � 0.0205 � 40025 ) = 411.4 �� ��� 420�� � = � + 70 = 420 + 70 , 490�� ��� 500��

���� = 18.5 = 700018.5 = 378.4�� ����� = 0.0205 � 300 � 420 = 2583 ��2 ��� �25�� ��� ���; = 491��2 ?�� 6�25 �� @: %���;� For negative moment = 202.5kN.m use the same section 300*420mm

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�� = � � � � � �2 � �� � (1 " 0.59 � � ���� �)202.5 � 106 = 0.9 � � 300 � 4202 � 400 � (1 " 0.59 � � 40025 )9.442 " + 0.01063 = 0 = 0.01204 ��� �� = 0.01204 � 300 � 420 = 1517.04��2Or ������ � � � 2 ��� 200��then �� = 202.5�1060.9�400(420"2002 ) = 1757.813��2 � = 1757.813�4000.85�25�300 = 110.294��

Second trial �� = 202.5�1060.9�400(420"110 .2942 ) = 1541.717��2 � = 1541.717�4000.85�25�300 = 96.735���� = 202.5�1060.9�400(420"96.7352 ) = 1513.592��2 � = 1513.592�4000.85�25�300 = 94.97��then �� = 202.5�1060.9�400(420"94.972 ) = 1510.006��2 � = 1510.006�4000.85�25�300 = 94.745 , 94.975�� :�or A� = �����2 = 202.5�1060.9�300�420�420 = 4.252 B = ��0.85��� = 4000.85�25 18.824 = 1B [1 " C1 " 2 � A� � B�� ] = 118.824 D1 " C1 " 2 � 4.252 � 18.824400 E = 0.01198As=0.01198*300*420= 1509.48mm2

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�� � �� = 0.85F�� � � � � + ��� � �� �������������������������������������������� "�%�&���'��(�)��������������� � �� = 0.85F�� � � � * � � + ��� � ���� � �� = 0.85F�� � * � 600�600"�� � � + ��� � �� bdfy �*�+���!&�,

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� = 0.85F�� � * � 600600"�� + ��� � ���� �*�+�� "

= 1B [1 " -1 " 2�A��B�� ] If � ��� then singly reinforced

Then ��1 = � � � � ���

� = ��1��0.85� �� ��� and ��1 = ���1 = 0.9 ��1 � �� (� " �2)

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Ex.6: Find the ultimate moment capacity for the section below, using fy= 345MPa and f’c=27.6MPa?

Assume fs = fs’= fy then As2 = As’

� = (�� " ��� )��0.85 � �� � � � = (4828 " 2414)3450.85 � 27.6 � 300 = 118��� = �* = 1180.85 = 139��

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��� = 600 (139 " 70)139 = 297.84 �G� < �� I = J1 + J���� � �� = 0.85 � �� � � * � � � � + 600(� " �� )� ���4828 � 345 = 0.85 � 27.6 � 0.85 � � � 300 + 600(� " 70)� � 2414�2 " 36.09 � " 16950 = 0C = 149.5mm then a = 127mm��� = 600 (149.5 " 70)149.5 = 319 �G� < �� :. �.

"��� = 0.75 � + � ����� = 0.75 � 0.85 � 0.85 � 27.6345 � 600600 + 345 + 2414300 � 530 319345= 0.0415�� = 4828300 � 530 = 0.0304 < "����� = � K0.85 � �� � � � � � !� " �2# + ��� � ��� (� " �� )L /106= 0.9 M0.85 � 27.6 � 300 � 127 &530 " 1272 ' + 2414 � 319(530 " 70)N /106= 694.08 kN.m

Ex. 7: Find the ultimate moment capacity and maximum live load that could be applied for the section of the beam below, using fy= 400 MPa and f’c=21MPa?

Try fs’=fs=fy

As2=As’=628mm2

� = (�� " ��� )��0.85 � �� � � � = (2945 " 628)4000.85 � 21 � 300= 173��� = �* = 1730.85 = 203.5����� = 600 (203.5"60)203.5 = 423.1 �G� > �� then the assumption is correct

"��� = 0.75 � + � = 0.75 � 0.85 � 0.85 � 21400 � 600600 + 400 + 628300 � 510 = 0.02117

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�� = 2945300 � 510 = 0.0192 < "����� = � K�� � �� !� " �2# + ��� � ��� (� " �� )L /106 = 0.9 K2316 � 400 !510 " 1732 # + 628 � 400(510 " 60)L /106= 454.8 kN.m

�� = �� � 22454.8 = �� � 422Wu= 56.65 kN/m

Wu = 1.2(0.3*0.6*24) + 1.6 WL = 56.65

Then WL = 32.17 kN/m

Ex.8: Find total area of steel required for the section below which supports Mu= 400kN.m , using fy=420MPa f’c=21MPaand d’= 65mm if you need compression reinforcement?

Try first as singly reinforced beam with ������ = 0.75 � 0.852 � 21420 � 600600 + 420 = 0.0159�� = � � � � � �2 � �� � (1 " 0.59 � � ���� �)= 0.9 � 0.0159 � 300 � 4752 � 420!1"0.59�0.0159�42021 #106 = 330.4��. � < 400��. �We need comp. reinf.

Mn1= 330.4kN.m

then Mn2= 400 - 330.4 = 69.6kN.m��1 = � � � � ��� = 300 � 475 � 0.0159 = 2265.75��2

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��2 = ��2���(� " �� ) = 69.6 � 1060.9 � 420(475 " 65) = 449��2As tension total= 2265.75+449= 2714.75 mm2

� = ��1 � ��0.85 � �� � � � = 2265.75 � 4200.85 � 21 � 300 = 177.7��� = �* = 177.70.85 = 209����� = 600 (209"65)209 = 413.4 �G� < �� = 420 �G� then

As’*fs’=As2*fy

��� = ��2 � ����� = 449 � 420423.4 = 456.2��2Check:

"��� = 0.75 � + � = 0.75 � 0.852 � 21420 � 600600 + 420 + 456.2300 � 475 413.4420 = 0.02023�� = 2265.75300 � 475 = 0.0159 < "���Ex.9 : Reinforced concrete rectangular beam with b=300mm and total depth = 600 mm itsupport service dead moment including its weight 200kN.m and service live moment 180kN.m, find area of steel required. using f’c=30MPa and fy= 400MPa?

To find d = h-100= 600-100= 500mm

Mu=1.2 Md +1.6 ML= 1.2*250 +1.6* 200= 620 kN.m

then find = 1B [1 " -1 " 2�A��B�� ]

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And ��� = 0.75 � = 0.75 � 0.852 � 30400 � 600600+400 = 0.0232 > ��� then we need comp. reinforced.

��1 = ���1 = 0.9 � 3480 � 400!500"1822 #106 = 512.39 kN.m

��2 = �� " ��1 = 620 " 512.39 = 107.61��. �Assume d’= 70mm

��2 = ��2��� (� " �� ) = 107.61 � 1060.9 � 400(500 " 70) = 695.16 ��2Ast= 3480 + 695.16 = 4175.16 mm2

� = �* = 1820.85 = 214mm

��� = 600 (214"70)214 = 403.74 �G� > �� = 400 �G� ��� = ��2 = 695.16 ��If using bars with �30mm then no. of bars required

� = 4175.16707 = 5.9 ��� ��� 6� 30�� for tension and for compression reinforcement 3 �20mm then As’= 3*314= 942 mm2.

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Analysis and design of T or L Beams:

���� � 4 � �@ + 16�� � O2

���� � 12 + �@ � �@ + 6�� � �@ + O2

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Analysis of T or L beams:

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Ex.10:find the moment section capacity for the (T) beam shown,using f’c= 20MPa and fy=400MPa :

1- If As=3000mm22- If As=4800mm2

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� = 0.85* ����� 600600 + �� = 0.852 20400 � 600600 + 400 = 0.0217

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Ex.11: A concrete Slab with 80mm supports on beams the distance between them 1.8m clc with simply supported span of 5m, find ultimate moment capacity for the interior beam, using f’c=20.7MPa and fy = 345MPa.Use :

1- 6 � 25 = 2950mm2 2- 8 � 32= 6436 mm2Solution:���� � 50004 = 1250��� 360 + 16 � 80 = 1640��� �� :� ����� = 1800��Then use ���� = 1250�� �� %����;For As = 2950mm2 try a = hf = 80mm

� = 2950�3450.85�20.7�1250 = 46.3 < 80�� then rectangular section

��� � �� � ��� ��� � ��� �4 �� � 1.4�� � �20.74 � 345 � 1.4345��� � 0.004�� = 2950600 � 360 = 0.01366��� = 0.75 � 0.852 � 20.7345 � 600600 + 345 = 0.0206Under reinforced beam

�� = 0.9 � 2950 � 345 &600 " 46.32 ' = 528.38 ��. �For As= 6436 mm2

� = 6436�3450.85�20.7�1250 = 100.9�� > 80�� then T-section

��� = 0.85 � �� �(���� " �@)���� = 0.85 � 20.7 � (1250 " 360) � 80400 = 3631��2

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���@ = (�� " ���)��0.85 � ��� � �@ = (6436 " 3631)3450.85 � 20.7 � 360 = 152.78����� � �� � @� ������ � ��� �4 �� � 1.4�� � �20.74 � 345 � 1.4345��� = 0.00329 � 0.004�� = 6436600 � 360 = 0.0297��� = 0.75P� + � Q = 0.75 � 0.852 � 20.7345 � 600600 + 345 + 3631360 � 600 = 0.0332

o.k. under reinforced.�� = �(�� + �@) = 0.9(��� � �� &� " ��2 ' + ��@ � �� !� " �2#)= 0.9 K3631 � 345 !600 " 802 # + 2805 � 345 !600 " 152.782 #L106 = 1087.4 ��. �

Ex.12:A simply supported beam(TT) shown below supports service dead load of 20kN/m and service live load of 35kN/m using strength design method find steel area required for the beam ,use f’c= 24MPa and fy= 400MPa d= 510 and d’=60mm?

Wd= (0.9*0.6 – 0.45*0.6) *24= 6.48kN/m

Wu= 1.2*(20+6.48) +1.6*35 =87.67kN/m

For –ve moment= -536.99kN.m��� = 0.75 �= 0.75 � 0.852 � 24400 � 600600 + 400 = 0.0195

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����� = � � ��� � � � �2 � �� � &1 " 0.59 � ��� � ���� �'�� = 0.9 � 0.0195 � 300 � 5102 � 400 � 1 " 0.59 � 0.0195 � 40024106 = 442.7��. �

< 536.99��. � RSTU VWT XYVZ\^ _T`UaY_bTcTUR.Mu2= 536.99 - 442.7 = 94.29 kN.m

As1= 0.0195*300*510=2983.5 mm2

��2 = 94.29 � 1060.9 � 400(510 " 60) = 582 ��2Ast= 2983.5 +582= 3565.5 mm2

� = 2983.5 � 4000.85 � 24 � 300 = 195��� = 1950.85 = 229.5����� = 600(229.5 " 60)229.5 = 443�G� > �� As’=As2= 582mm2

Check: ��� � �� � ���" = 0.75� + ���� � ��� �4 �� � 1.4�� � �20.74 � 345 � 1.4345��� = 0.00329 � d. ddeFor T or L (ACI-Code - section10.5.2 — For statically determinate members with aflange in tension, As,min shall not be less than the value given by Eq. (10-3)

except that bw is replaced by either 2bw or the width of the flange, whichever is smaller).

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�� = ���@ � ��� = 3565.5300 � 510�� = 0.0233���" = 0.75 � 0.852 � 24400 � 600600 + 400 + 582300 � 510 = 0.0195 + 0.0038 = 0.0233O.K. under reinforced beam.

For +ve moment= 639.58kN.m

Try a =hf = 150mm

Mu= 0.9*0.85* f’c* beff* hf ( d - hf/2)

= 0.9*0.85*24*900*150 (510- 150/2)/106 =1078.15 kN.m >639.58kN.m

then rectangular section.

A� = �����2 = 639.58 � 1060.9 � 900 � 5102 = 3.036 B = ��0.85 ��� = 4000.85 � 24 = 19.61 = 1B D 1 " C1 " 2 � A� � B�� E = 119.61 D 1 " C1 " 2 � 3.036 � 19.61400 E = 0.0082As= 0.0082* 300*510=1254.6mm2 ��� � �� � ��� = 0.75�0.004 � 0.0082 � 0.0195 o.k.

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Ex.13: Find area of steel required for the beams B-1and B-2 for the plan shown, using f’c = 20MPa and fy=400MPa.

WL 100 50 kN/m

Loads B-1 B-2

Wd 55 27.5 kN/m (including its own weight)

For interior beam B-1: ���� � 50004 = 1250�� � 250 + 16 � 75 = 1450�� � �/� :� ����� = 3000��Then use ���� = 1250�� �� %����;Wu= 1.2 *55 + 1.6 * 100= 226 kN/m

Mu= 226*(5)2

Try a = hf = 75mm

�� = � � 0.85 �� � � ���� � �� &� " ��2 '�� = 0.9 � 0.85 � 20 � 1250 � 75!500"752 #106 = 663.4 < 706.3��. � then T- section

��� = 0.85 � �� �(���� " �@)���� = 0.85 � 20 � (1250 " 250) � 75400 = 3188����� = � � ��� � �� &� " ��2 ' = 0.9 � 3188 � 400 !500 " 752 #106 = 530.8 ��. �Muw= 706.3 - 530.8 = 175.5 kN.m

��@ = � � 0.85 �� � � �@ � � !� " �2#175.5 � 106 = 0.9 � 0.85 � 20 � 250 � � !500 " �2#

/8 = 706.3 kN.m

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�2 " 1000� + 91764.7 = 0a = 102.2 mm

��@ = 175.5 � 1060.9 � 400(500 " 102.22 ) = 1085.9��2

As-total = 3188+1085.9= 4273.9mm2

Check ��� � �� � ��� ��� � ��� �4 �� � 1.4�� � �204 � 400 � 1.4400��� = 0.0028 � d. ddfg�� = 4273.9500 � 250 = 0.0342��� = 0.75P� + � Q = 0.75(0.852 � 20400 � 600600 + 400 + 3188500 � 250) = 0.0503Under reinforced beam

For exterior beam B-2: ���� � 500012 + 250 = 667�� � 250 + 6 � 75 = 700 �� � (�@ + �� :� �����) 2 = 250 + 30002 = 1625 ��Then use ���� = 667 �� �� %����;Wu= 1.2 *27.5 + 1.6 * 50= 113 kN/m

Mu= 113*(5)2/8 = 353 kN.m

Try a = hf = 75mm

�� = � � 0.85 �� � � ���� � �� &� " ��2 '

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�� = 0.9 � 0.85 � 20 � 667 � 75 !500 " 752 #106 = 353.98 h 353.12��. � Then rectangular- section

A� = �����2 = 353.12 � 1060.9 � 667 � 6002 = 2.353 B = ��0.85 ��� = 4000.85 � 20 = 23.53 = 1B D 1 " C1 " 2 � A� � B�� E = 123.53 D 1 " C1 " 2 � 2.353 � 23.53400 E = 0.0064As= 0.0064* 250*500=800 mm2 ��� � �� � ��� = 0.75�0.0035 � 0.0064 � 0.03114 o.k.

Analysis and design of Irregular sections ���� ����������:

�-�&�����(�.�!(Mn )�����-/���0����1�)���2�34"5��678������!���.�!��'��#����"�/�9�:#!"��:-�&���;��#��<=��9>8��.�!�2�34"?�����!��>�7���@0�8#�iF� = 0 J = I 0.85 �� � � ��:�$ = �� � ��

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Then ��:�$ = jW�a^d.kg a�b�-����#"��&�� (���A��)����!���B=��.�!���#���CD��9��8������0�E��@���FGAcomp��H�G�B5+���,�!���

��G�!������/�.�!������5������-/���I�$�<�����!���7� C(����#��73J�����!���K"��X�7 ���K"����H���.*�+5�;8G��<����7�3 (����#�

�.�!��L7�� ���#��<�#!��<��<��?��M�N��-�&��������H��������X��.�!���8,�b1) (O�N�N���;,�D(�)��

����!���B=��>�7��-/���.�!���#���1 �" = iA�yiA �� = m�� � ��(� " �")���������!��.�!�����G������P�4$�.�!����Asb �.�!���1��Asmax

�� = 600�600 + �� ��� �� = * � ���Q�8���=�,�2�34"?�����!���!"��� �� � � �� = 0.85 �� � � ��:�$�

�� � = 0.85 �� ����:�$ ��� and �� ��� = 0.75 �� � � �� �� Ex.14 : Find the moment section capacity for the beam shown, using f’c = 30MPa and fy = 400 MPa and As=2800mm2?

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0.85 �� � � ��:�$ = �� � �� Then ��:�$ = 2800�400 0.85�30 = 43921 ��2 Area of equal = 0.5*300*150 = 22500mm

�� = 4 � (28)24 = 2464��20.85 �� � � ��:�$ = �� � �� Then ��:�$ = 2464�400 0.85�25 = 46381.2 ��2

2

so we must add another area X*b

43921= 22500+300*X

then X = 71 mm and a = 150+71= 221 mm � = 2210.85 = 260��Calculate�� = 600�600+�� = 600�650600+400 = 390�� > � ��� ����; ;����. ���� �%�: �����:�� = * � �� = 0.85 � 390 = 331.5����:�$� = 0.5 � 300 � 150 + (331.5 " 150) � 300 = 76950��2�� � = 0.85 �� ����:�$ ��� = 0.85�30�76950400 = 4905.56��2�� ��� = 0.75 � 4905.56 = 3679.17��2 � �� �� = 2800��2�� = J1 (650 " 100) + J2( 500 " �2)�� = 0.9 � 0.85 � 30[1502 (550) + 300 � 71 & 500 " 712 ' 10"6 = 511.2 ��. �Ex.15: Find ultimate section capacity for the section, using f’c =25 MPa and fy =400MPa.

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Area of equal = 100(200+400)2 = 30000mm2 so we must add another area X*b46381.2 = 30000 + 400*X then X = 40.95 mm and a = 100+40.95= 140.95 mm� = 140.950.85 = 165.8��Calculate �� = 600�600+�� = 600�400600+400 = 240�� > � ��� ����; ;����. ���� �%�: �����:�� = * � �� = 0.85 � 240 = 204����:�$� = 30000 + (204 " 100) � 400 = 71600��2�� � = 0.85 �� ����:�$ ��� = 0.85�25�71600400 = 3803.75��2�� ��� = 0.75 � 3803.75 = 2852.81��2 � �� �� = 2464��2�" = 400 � 140.9122 " 1003 � 2 100 � 100246381.2 = 78.41�� �� = m�� � ��(� " �") = 0.9 � 2464 � 400 (400 " 78.41)106 = 285.26��. �Ex. 16: Find area of steel required for the beams shown, using f’c = 30MPa and fy = 400MPa, the external moment applied is Mu= 450kN.m?

Try a= 150mm

�� = 0.9 � 0.85 � 30 � 200 � 150 !530 " 1502 #106 = 313.27��. �< 450��. �Add part of web: 450-313.27=136.73kN.m

136.73*10P

6P= 0.9* 0.85 * 30 [400 * x (530-150-x/2)]�2 " 760� + 29788.67 = 0

� = 760 ± �7602 " 4 � 29788.672 = 41.46���� � �� = 0.85�� � � ��:�$�� � 400 = 0.85 � 30 � [200 � 150 + 41.46 � 400]Ast = 2969.73mmP

2P

�� = 600�600+�� = 600�530600+400 = 318���� = * � �� = 0.85 � 318 = 270.3��

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�� ��� = 0.75 �� � = 0.75 0.85 �� � � ��:�$���= 0.75 0.85 � 30 � [150 � 200 + 400 � 120.3]400 = 3735.113��2> 2969.73��2 ��� ����; ;����:;��� ����.Ex.17: Using the method of irregular section repeat example 11, using f’c=20.7MPa and fy = 345MPa , As= 6436mm2,

0.85 �� � � ��:�$ = �� � ��

beff = 1250mm and d=600mm?

Then ��:�$ = 6436�345 0.85�20.7 = 126196.07 ��2 Area of flange = 80*1250 = 100000 mm2

so we must add another area X*bw

126196.07 = 100000 + 360*X then X = 72.767mm

� = 1250 � 80 � 40 + 360 � 72.767(80 + 72.767)2126196.07 = 55.86 ��Calculate �� = 600�600+�� = 600�600600+345 = 380.95���� = * � �� = 0.85 � 336.5 = 323.808����:�$� = 100000 + (323.808 " 80) � 360 = 187770.88 ��2�� � = 0.85 �� ����:�$ ��� = 0.85�20.7�187770 .88345 = 9576.3��2�� ��� = 0.75 � 9576.3 = 8139.87��2 � �� �� = 6436��2�� = m�� � ��(� " �") = 0.9 � 6436 � 345 (600 " 55.86 )106 = 1087.4��.

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Sheet no. 2

Q.1 : Check the adequacy of the beam shown using f’c=34MPa and fy=380MPa?

Q.2: Find maximum load that could be applied on the beam shown , using f’c=30MPa and fy=400MPa and As=2000mm2,the ratio of live load to dead load =2 (neglect weight of beam)?

Q.3: determine the maximum live load that could be applied if A’s=0.35 As, f’s= 0.89 fy , fy=400MPa and f’c=30MPa. (Neglect weight of beam)?

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Q.4: For the section below find ultimate section capacity if As=3800mm2 for the first two ,f’c=25 and fy=380MPa?

Q.5: Design the beam shown, it support dead load (including its own weight ) =10kN/mand one concentrated live load of 30kN. Use f’c=24MPa and fy =380MPa?

Q.6: Find total area of steel required for (positive and negative moment) the beam shown, the material properties are f’c=25MPa and fy= 400Mpa?