Tgs Baja
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1. Sebuah batang tarik berukuran 10 mm x 175 mm disambung dengan 3 Buah baut berdiameter 25 mm. Mutu baja yang digunakan adalah BJ 37. hitunglah tahanan tarik rencana batang tersebut dengan mengasumsikan A e = A n` 10
O
O
mm
x
175
mm
O
JAWAB : Luas kotor, Ag Lebar lubang An
= = =
10 25 Ag
x + −
=
1750
−
2 175 = 1750 mm 6 = 31 mm ( lebar lubang x tebal plat )
10 31
=
2 1440 mm
<
85% Ag
=
2 1488 mm
2. Sebuah batang tarik dari pelat berukuran 10 mm x 190 mm, harus memikul beban mati sebesar 110 kN dan beban hidup 200 kN mutu baja BJ 41 dan diameter baut 25 mm Dengan mengasumsikan A e = A n` periksalah kecukupan batang tersebut ! 10
Ο Ο
mm
x
190 mm
Ο Ο
JAWAB : Menghitung Luas netto, A n : Pot. 1-2-3 : An
=
10
320
-
3
25
+
2
=
2390
mm2
ο1 ο4 6
ο2
ο
4 @ 60
ο
320
5
ο3 50
80 1850
11733.33
Pot. 1-4-2-5-3 : An
=
10
320
-
5
25
+
2
+
4
2 80 ` x 4 x
110 60
=
13583.3 mm2
320
-
4
25
+
2
+
4
2 80 ` x 4 x
110 60
=
13853.3 mm2
Pot. 1-4-5-3 : An
=
10
Pot. 1-4-6 : An
=
10
=
15269.2
85% A g Jadi, A n
=
320
-
3
=
Koefisien reduksi U
+
2
=
29920
2390 = =
mm2 1 - x/L 0.962 >
2 80 ` x 4 x
110 60
+
mm2
⅟2
1 →
U
=
138.24
=
0.90
6400
240
Kondisi Fraktur : Ae = U.A n
=
0.9
2390
=
Φ Tn
=
0.75
2151
370
Φ Ae.fu
4
= 0.9
Kondisi leleh : Φ Tn = Φ Ag.fy
=
+
mm2
0.85 320 110
min
25
2151 =
=
10 `/ 0.9 ton
mm2
59.6903 ton
130
2 50 ` x 4 x
110 60
O
O
mm
x
175
mm
O
JAWAB : Luas kotor, Ag Lebar lubang An
= = =
10 25 Ag
x + −
=
1750
−
2 175 = 1750 mm 6 = 31 mm ( lebar lubang x tebal plat )
10 31
=
2 1440 mm
<
85% Ag
=
2 1488 mm
2. Sebuah batang tarik dari pelat berukuran 10 mm x 190 mm, harus memikul beban mati sebesar 110 kN dan beban hidup 200 kN mutu baja BJ 41 dan diameter baut 25 mm Dengan mengasumsikan A e = A n` periksalah kecukupan batang tersebut ! 10
Ο Ο
mm
x
190 mm
Ο Ο
JAWAB : Menghitung Luas netto, A n : Pot. 1-2-3 : An
=
10
320
-
3
25
+
2
=
2390
mm2
ο1 ο4 6
ο2
ο
4 @ 60
ο
320
5
ο3 50
80 1850
11733.33
Pot. 1-4-2-5-3 : An
=
10
320
-
5
25
+
2
+
4
2 80 ` x 4 x
110 60
=
13583.3 mm2
320
-
4
25
+
2
+
4
2 80 ` x 4 x
110 60
=
13853.3 mm2
Pot. 1-4-5-3 : An
=
10
Pot. 1-4-6 : An
=
10
=
15269.2
85% A g Jadi, A n
=
320
-
3
=
Koefisien reduksi U
+
2
=
29920
2390 = =
mm2 1 - x/L 0.962 >
2 80 ` x 4 x
110 60
+
mm2
⅟2
1 →
U
=
138.24
=
0.90
6400
240
Kondisi Fraktur : Ae = U.A n
=
0.9
2390
=
Φ Tn
=
0.75
2151
370
Φ Ae.fu
4
= 0.9
Kondisi leleh : Φ Tn = Φ Ag.fy
=
+
mm2
0.85 320 110
min
25
2151 =
=
10 `/ 0.9 ton
mm2
59.6903 ton
130
2 50 ` x 4 x
110 60
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