1

Technische Mechanik - 1 WS 2015/16

1. Aufgabe

∑F →= 0↝ AX = 0

∑F ↑= 0↝ AY = 3G = 300N

↷ ∑MA = 0↝MA − 400Nm

(I) ∑F →= 0 = Q (X1)

∑F ↑= 0↝ N (X1) = −300N

↶ ∑Ms = 0↝M (X1) = −400Nm

(II) ∑F ↗= 0↝ N (ϕ = 0○) = −200 kN ; N (ϕ = 90○) = 0

∑F ↘= 0↝ Q (ϕ = 0○) = 0; N (ϕ = 90○) = 200 kN

↶ ∑Ms = 0↝M (ϕ = 0○) = −400Nm; M (ϕ = 90○) = 200Nm;

(III) ∑F →= 0 = N (X3)

∑F ↑= 0↝ Q (X3) = 100N

Institut fur Angewandte Mechanik 1

Technische Mechanik - 1 WS 2015/16

↶ ∑Ms = 0↝M (X3 = 0) = 0; M (X3 = 2a) = −200Nm

Institut fur Angewandte Mechanik 2

Technische Mechanik - 1 WS 2015/16

2. Aufgabe

∑F → = 0

BX +CX = 0 (1)

CY − F = 0 (2)

AZ +BZ +CZ +R = 0 (3)

∑MC ↷ = 0

2aBZ + aAZ − aF + 4Fa = 0 (4)

−2AZ − 4aF = 0

↝ AZ = −2F

−2aBX − 2aF = 0

↝ BX = −F

AZ in (4)↝ BZ = −F /2

BX in (1)↝ CX = F

AZ ,BZ in (3)↝ CZ = −3F /2

Institut fur Angewandte Mechanik 3

Technische Mechanik - 1 WS 2015/16

3. Aufgabe

XS = ∑XiAi

∑Ai

YS = ∑YiAi

∑Ai

i Ai Xi Yi XiAi YiAi

I 16a2 4a 5a+4a

364a3 101.33a3

II 40a2 4a5a

2160a3 100a3

III -6a2 2a3a

2-12a3 -9a3

∑ 50a2 212a3 192.33a3

XS = 4.24a

YS = 3.85a

Institut fur Angewandte Mechanik 4

Technische Mechanik - 1 WS 2015/16

4. Aufgabe

∑MB ↶ = 0

↝ AY = −F

∑F → = 0

↝ BX = 0

∑F ↑ = 0

↝ BY = 2F

∑F → = 0

↝ S1 = S2

3(1)

∑F ↑ = 0

↝ S2 = −2F

√3

2

S2 in (1)↝ S1 = −F

∆l1 = −Fl1EA

∆l2 = −√

3Fl2EA

∆X =√

3

2

F

EA(l1 − l2)

∆Y = F

EA(3

2l2 + 1

2l1)

Institut fur Angewandte Mechanik 5

Technische Mechanik - 1 WS 2015/16

6. Aufgabe

(a) σY = 100N/mm2

εX = 0 (Starre Wandung)

↝ σX = −30N/mm2

(b) Langeanderung in y-Richtung

εY = −0.0013

∆Y = 0.013 cm

Institut fur Angewandte Mechanik 6