H = 0kNA V = 9kNA V = 9kNB q =12kN/m1 q =12kN/m 1,5 1,5 1,5 1,5
N
[kN]
9 9 9+
+
-T
[kN]
+
-9 9M
[kNm]
styczna pozioma styczna pozioma
styczna pozioma
Belki proste z obciążeniem trójkątnym
Zadanie 1. Narysuj wykresy sił wewnętrznych T, N, M dla poniższych układów. a) Wyznaczenie reakcji: kN V V MB A A 9 6 27 81 0 6 5 , 1 3 12 2 1 5 , 4 3 12 2 1⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ − ⋅ = ⇒ = − = =
∑
kN V V MA B B 9 6 27 81 0 6 5 , 1 3 12 2 1 5 , 4 3 12 2 1 = − = ⇒ = ⋅ − ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ =∑
0 = =∑
Rx HA Sprawdzenie: 0 9 3 12 5 , 0 3 12 5 , 0 9− ⋅ ⋅ + ⋅ ⋅ − = =∑
Rystrona 2 H = 0kNA V = 18kNA V = 18kNB q =12kN/m1 q =12kN/m
1,5
1,5
1,5
1,5
N
[kN]
18 9 9+
T
[kN]
22,5M
[kNm]
18 -styczna pozioma styczna pozioma styczna pozioma 22,5 27+
b) Wyznaczenie reakcji: kN V V MB A A 18 6 27 81 0 6 5 , 1 3 12 2 1 5 , 4 3 12 2 1⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ = ⇒ = + = =∑
kN V V MA B B 18 6 27 81 0 6 5 , 1 3 12 2 1 5 , 4 3 12 2 1⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ = ⇒ = + = =∑
0 = =∑
Rx HA Sprawdzenie: 0 18 3 12 5 , 0 3 12 5 , 0 18− ⋅ ⋅ − ⋅ ⋅ + = =∑
Ryq =18kN/m1 q =18kN/m2 H = 0kNA V = 5,4kNA V = 5,4kNB 2 3 3 2
N
[kN]
T
[kN]
styczna pozioma styczna pozioma
5,4 5,4 5,4 5,4 21,6 + -+ 10,8 15,63 15,63 10,8 -+
M
[kNm]
e e1 c) Wyznaczenie reakcji: kN V V MB A A 5,4 10 108 162 0 10 6 3 18 2 1 4 3 18 2 1⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ = ⇒ = − = =∑
kN V V MA B B 5,4 10 108 162 0 10 6 3 18 2 1 4 3 18 2 1⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ + ⋅ = ⇒ = − = =∑
0 = =∑
Rx HA Sprawdzenie: 0 4 , 5 3 18 5 , 0 3 18 5 , 0 4 , 5 − ⋅ ⋅ + ⋅ ⋅ − = =∑
RyWyznaczenie ekstremum: z proporcji:
x x x q x x q 6 3 18 ) ( ) ( 3 18= ⇒ = = 2 = − = ⋅ ⋅ − = ⋅ − = q =18kN/m x q(x)
strona 4 P=15kN q =6kN/m q =12kN/m1 H = 0kNA V = 15,43kNA V = 41,57kNB
3
4
2
N
[kN]
styczna pozioma 15,43 2,57 26,57 15 15 + +-T
[kN]
30 28,29 28,57 +-M
[kNm]
e 3 ) 2 ( 4 , 5 3 6 5 , 0 ) 2 ( 4 , 5 3 ) ( 5 , 0 ) 2 ( 4 , 5 ] [x x q x x x x x x x x x M = ⋅ + − ⋅ ⋅ ⋅ = + − ⋅ ⋅ ⋅ = + − kNm m x M[ =1,34 ]=5,4(1,34+2)−(1,34)3 =15,63 d) Wyznaczenie reakcji: kN V V MB A A 15,43 7 5 18 48 30 0 7 ) 1 4 ( 3 12 2 1 2 4 6 2 15⋅ − ⋅ ⋅ − ⋅ ⋅ ⋅ + + ⋅ = ⇒ =− + + ⋅ = =∑
kN V V MA B B 41,57 7 9 15 5 24 36 0 7 9 15 ) 2 3 ( 4 6 2 3 12 2 1⋅ ⋅ ⋅ + ⋅ ⋅ + + ⋅ − ⋅ = ⇒ = + ⋅ + ⋅ = =∑
0 = =∑
Rx HA Sprawdzenie: 0 57 , 41 15 4 6 3 12 5 , 0 43 , 15 − ⋅ ⋅ − ⋅ − + = =∑
Ryq =12kN/m 3 x q(x) 1 Wyznaczenie ekstremum: z proporcji: x x x q x x q 4 3 12 ) ( ) ( 3 12= ⇒ = = 0 2 43 , 15 4 5 , 0 43 , 15 ) ( 5 , 0 43 , 15 ] [x = − q x ⋅x= − ⋅ x⋅x= − x2 = T m x x 78 , 2 0 2 43 , 15 2 = ⇓ = − 3 3 2 43 , 15 3 4 5 , 0 43 , 15 3 ) ( 5 , 0 43 , 15 ] [x x q x x x x x x x x x M = ⋅ − ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ = − kNm m x M (2,78) 28,57 3 2 78 , 2 43 , 15 ] 78 , 2 [ = = ⋅ − 3=