4. Kombinasi Lentur dan geser
�
�
∅�
�
+ 0.625 �
�
∅�
�
≤ 1.35 0.895+ 0.625 . 0.586
≤ 1.35
1.26 1.35 …….ok
B. Desain Bresing
Gaya-gaya dalam maksimum yang bekerja Nu = 146.78 kN ; 1.25 . 1.5 Nu = 275.213 kN
Mu = 30.567 kNm ; 1.25 . 1.5 Mu = 57.30 kNm Bresing didesain berdasarkan kekuatan link:
Nu = 1.25 . 1.5 Vn link = 1.25 . 1.5 . 157.25 = 294.84 kN Mu = 1.25 Ry.0.5Vn.e = 58.97 kNm 1.25 . 1.5 Mu
Maka diambil Nu = 294.84 kN
Mu = 58.97 kNm Coba penampang WF 200.150.6.9
1. Cek Kelangsingan Penampang
Flange: �
�
=
� 2
�
�
=
150 2.9
= 8.33 �
�
=
170 ��
�
=
170 √240
= 10.97 λ
f
≤ λ
p
penampang kompak
Universitas Sumatera Utara
Web: �
�
=
ℎ �
�
=
200 −2.9
6
= 30.33 �
�
∅�
�
= �
�
∅�
�
. �
�
= 294.84
0.85 . 3901 . 240 = 0.3705
≥ 0.125
Maka, �
�
= ���. �
500 ��
�
�2.33 −
�
�
∅�
�
� ,
665 ��
�
� = 70.23; 42.92 = 70.23 λ
w
≤ λ
p
2. Analisa komponen tekan
penampang kompak
kcx = kcy = 1 Analisa tekuk bresing
L = 5.97 m = 5970 mm fy = 240 MPa
E = 200000 MPa Arah x:
Lkx = 5970 mm �
�
= �
��
�
�
= 5970
83 = 71.93
≤ 200 ��‼
�
��
= 1
� �
�
��
�
� =
1 �
. 71.93. �
240 200000
= 0.793 0.25
≤ λ
cx
≤ 1.25 � =
1.43 1.6
− 0.67 �
��
= 1.43
1.6 − 0.67 . 0.793
= 1.34
�
��
= �
�
. �
�
� = 3901.
240 1.34
= 698686.6 � = 698.687 ��
Universitas Sumatera Utara
Arah y: Lky = 5970 mm
�
�
=
�
��
�
�
=
5970 67 .6
= 88.31 200 ok
�
��
= 1
� .
�
�
. ��
�
� =
1 �
. 88.31. �
240 200000
= 0.974 0.25
≤ λ
cy
Nn = min Nnx; Nny = Nny = 620.03 kN ≤ 1.25
� = 1.43
1.6 − 0.67�
��
= 1.43
1.6 − 0.67.0.974
= 1.51
�
��
= �
�
. �
�
� = 3901.
240 1.51
= 620026.5 � = 620.03 ��
ØNn = 0.85 . 620.03 = 527.03 kN ØNn
≥ Nu Nu = 294.84
Strength rasio = 294.84527.03 = 0.56 ok
3. Kapasitas Lentur
Analisa tahanan lentur nominal bresing Lb = 5970 mm
�
�
= 1.76 . �
�
. �
� �
�
= 1.76 . 36.1 . �
200000 240
= 1834 ��
�
�
= �
�
− �
�
= 240 − 0.3 . 240 = 168 ���
Universitas Sumatera Utara
� = �
21 − ʋ
= 200000
2 1 + 0.3 = 76923.1
���
� = 1
3 .
�ℎ − 2. �
�
��
� 3
+ 2. 1
3 .
�. �
� 3
= 1
3 . 200
− 2.9. 6
3
+ 2. 1
3 . 150. 9
3
= 86004 ��
4
�
1
= �
�
�
. ��
� � � 2
= �
277000 .
� 200000 . 76923.1 . 86004 . 3901
2 = 18219.89
��� �
�
= �
�
. ℎ − 2�
� 2
4 = 5070000.
200 − 2 . 9
2
4 = 4.2
� 10
10
�
2
= 4 �
�
�
� �� .
�
�
�
�
= 4 �
277000 76923.1 . 86004
� . 4.2
� 10
10
5070000 = 1.387
� 10
−4
�
�
= �
�
� �
1
�
�
��1 + �1 + �
2
. �
� 2
= 36.1 �
18219.89 168
� �1 + �1 + 1.387 � 10
−4
. 168
2
Lr = 7022 mm
Lp ≤ LbLr
�
�
= �
�
��
�
− ��
�
− �
�
� � �
�
− �
�
�
�
− �
�
��
Ma = momen di ¼ bentang = 10.52 kNm Mb = momen di ½ bentang = 52.76 kNm
Mc = momen di ¾ bentang = 50.48 kNm �
�
= 12.5
�
��� .
2.5 �
���
+ 3 �
�
+ 4 �
�
+ 3 �
�
= 1.39 �
�
= �
�
. �
�
= 277000 . 168 = 46536000 ��� = 46.536 ���
Universitas Sumatera Utara
�
�
= 1.12 �
�
. �
�
= 1.12 . 277000 . 168 = 51120320 ��� = 51.12 ���
�
�
= 1.39 �51.12 − 51.12 − 46.536 �
5970 − 1834
7022 − 1834�
� = 65.98 ���
ØMn = 0.9 x 65.98 = 59.38 kNm Mu = 58.97 kNm
ØMn Mu ok Strength rasio = 58.9759.38 = 0.99 ok
4. Persamaan interaksi aksial momen Mn hanya arah x