1 1 Lösungen zum Arbeitsblatt Nr.1 1.1 Aufgabe 1 S235(St37) : HEA140 : fyk = 23, 5 kN/cm2 fuk = 36, 0 kN/cm2 A = 31, 4 cm2 Anet = 31, 4 − 4 · 0, 85 · 1, 8 Npl,Rd = Nu,Rd 1.2 1.2.1 = = A · fyk 31, 4 · 23, 5 = γm0 1, 0 0, 9 · Anet · fuk 0, 9 · 25, 3 · 36, 0 = γm2 1, 25 25, 3 cm2 = 737, 9 kN = 655, 8 kN (maßgebend) Aufgabe 2 Stabnachweis fy = 23, 5 kN/cm2 fu = 36, 0 kN/cm2 NEd = 1, 35 · 55 L 7 0 x 7 A * 3 0 S235(St37) : 1 8 fuk = A∗ 1, 8 ) · 0, 7 = 1, 47 cm2 2 36, 0 = 84, 7 kN = 2, 94 · 1, 25 74 = = 0, 87 < 1 84, 7 NEd Nt,Rd 4 0 γM2 = (3, 0 − 7 7 0 Nachweis des Knotenbleches 23, 5 kN/cm2 fuk = 36, 0 kN/cm2 A = 2 · 5, 3 · 0, 6 = 6, 36 cm2 Anet = 6, 36 − 0, 6 · 1, 8 = 5, 28 cm2 Npl,Rd = = NEd Nt,Rd = B r u c h lin ie l = 5 5 .c o s 1 5 ° = 149, 5 kN = 136, 9 kN = 0, 54 < 1 1 5 ° Nu,Rd 6, 36 · 23, 5 1, 0 0, 9 · 5, 28 · 36, 0 1, 25 74 136, 9 1 5 ° 1 5 = 4 0 fyk 4 0 S235(St37) : ⇒ 74 kN 4 0 + 5 5 .t a n 1 5 ° = 5 5 1.2.2 = Nu,Rd Nu,Rd ⇒ 2 · A∗ · 5 5 2 Aufgabe 3 1.3.1 Stabnachweis S235(St37) : 2 × U160 : fyk = 23, 5 kN/cm2 fuk = 36, 0 kN/cm2 NEd = A = 2 · 24, 0 = 48, 0 cm2 Anet = 48, 0 − 4 · 0, 75 · 2, 2 = 41, 4 cm2 Npl,Rd = ⇒ NEd Nt,Rd = = 1.128, 0 kN = 1073, 0 kN = 0, 84 < 1 Nachweis des Knotenbleches im Schnitt I-I 900 3 NEd = A = (2 · 6, 6 + 7, 0) · 1, 5 = 30, 3 cm2 Anet = 30, 3 − 2 · 1, 5 · 2, 2 23, 7 cm2 = B r u c. h l i n i e I - I l= 7 0 c o s 2 0 ° = 6 6 4 5 = = 614, 3 kN l = 1 2 9 .c o s 2 0 ° = 1 2 1 0, 49 < 1 3 5 NEd Nt,Rd = 712, 1 kN B r u c h lin ie III- III 3 5 = = 7 0 Nu,Rd 30, 3 · 23, 5 1, 0 0, 9 · 23, 7 · 36, 0 1, 25 300 614, 3 300 kN 4 5 + 7 0 .t a n 2 0 ° = Npl,Rd = ⇒ = 48, 0 · 23, 5 1, 0 0, 9 · 41, 4 · 36, 0 1, 25 900 1073 1 2 9 1.3.2 Nu,Rd 900 kN 4 5 + 2 3 0 .t a n 2 0 ° = 1.3 Nachweis des Knotenbleches im Schnitt III-III 4 5 1.3.3 NEd = A = (2 · 12, 1 + 7, 0) · 1, 5 = 46, 8 cm2 Anet = 46, 8 − 2 · 1, 5 · 2, 2 40, 2 cm2 ⇒ Nu,Rd = NEd Nt,Rd = 46, 8 · 23, 5 1, 0 0, 9 · 40, 2 · 36, 0 1, 25 900 1042 = = 1.099, 8 kN = 1.042, 0 kN = 0, 86 < 1 2 0 ° Npl,Rd = 900 kN 8 0 8 0 7 0
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