Cek Kelangsingan Penampang Analisa komponen tekan Kapasitas Lentur

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