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РАСЧЕТНО – ГРАФИЧЕСКАЯ РАБОТА № 1

по дисциплине: «Сопротивление материалов»

на тему «Построение эпюр внутренних усилий»

шифр 876 408

Задача 1

a = 6             q₁ = 20 кН          P₂ = 80 кН

b = 4,7          q₄ = 10 кН          P₄ = 90 кН

c = 4

d = 3

1) 0 ≤ x₁ ≤ d

Nx₁ = q₁∙x₁ 

x₁ = 0;             N_x1 = 20∙0 = 0

x₁ = d;             N_x1 = 20∙3 = 60

2) d ≤ x₂ ≤ c + d

N_x2 = q₁d – P₂ = -20 кН

3) c + d ≤ x₃ ≤ b + c + d 

N_x3 = q₁d – P₂ = -20 кН

4) b + c + d ≤  x₄ ≤ a + b + c + d

N_x4 = q₁d – P₂ + P₄ + q₄(x₄ – b – c – d)

x₄ = b + c + d;            N_x4 = -20 + 90 + 10∙0 = 70 кН

x₄ = a + b + c + d;     N_x4 = -20 + 90 + 10∙6 = 130 кН

Задача 2

0,1a = 0,6     q₁ = 20 кН          P₂ = 80 кН

0,1b = 0,47   q₄ = 10 кН          P₄ = 90 кН

0,1c = 0,4     M₂ = 60 кНм

0,1d = 0,3     M₃ = 110 кНм

1) 0 ≤ x₁ ≤ 0,1a

Qx₁ = -q₁∙x₁

M_x1 = -q₁∙x₁∙x₁/2

x₁ = 0;             Q_x1 = -20∙0 = 0 кН

                        M_x1 = 0 кНм

x₁ = 0,1a;        Q_x1 = -20∙0,6 = -12 кН

                        M_x1 = -20∙0,6∙0,6∙0,5 = 3,6 кНм

2) 0,1a ≤ x₂ ≤ 0,1(a+b)

Q_x2 = -q₁∙0,1∙a – P₂ = -92 кН

M_x2 = -q₁∙0,1a(x₂ - 0,1a/2)+M₂-P₂∙(x₂-0,1a)

x₂ = 0,1a;        M_x2 = -20∙0,6(0,6-0,6/2)+60-80∙(0,6-0,6)=56,4 кНм

x₂ = 0,1(a+b)  M_x2 = -20∙0,6(1,07-0,6/2)+60-80∙(1,07-0,6) = 13,16 кНм

3) 0,1(a+b) ≤ x₃ ≤ 0,1(a + b + c) 

Q_x3 = -q₁∙0,1∙a – P₂ = -92 кН 

M_x3 = -q₁∙0,1a(x₃ - 0,1a/2)+M₂-P₂∙(x₃-0,1a)+M₃

x₃ = 0,1(a+b);             M_x3 = -20∙0,6(1,07-0,6/2)+60-80∙(1,07-                     0,6)+110 = 123,16 кНм

x₃ = 0,1(a+b+с);         M_x3 = -20∙0,6∙(1,47-0,6/2)+60-80∙(1,47-0,6)+110 = 86,36 кНм

4) 0,1(a + b + c + d) ≤  x₄ ≤ 0,1(a + b + c + d) 

Q_x4 = -q₁∙0,1∙a – P₂ - P₄ - q₄∙(x₄ – 0,1∙(a + b + c))

M_x4 = -q₁∙0,1a∙(x₄ - 0,1a/2)+M₂-P₂∙(x₄-0,1a)+M₃-P₄∙(x₄-0,1∙(a+b+c))-q₄∙(x₄-0,1∙(a+b+c))²/2

x₄ = 0,1(a + b + c);                 Q_x4 = -90 - 92 = -182 кН

M_x4 = -20∙0,6∙(1,47-0,6/2)+60-80∙(1,47-0,6)+110 – 90∙(1,47-1,47) – 10∙(1,47-1,47)²/2=86,36 кНм

x₄ = 0,1(a + b + c + d);         Q_x4 = -182-10∙0,3 = -185кН

M_x4 = -20∙0,6∙(1,77-0,6/2)+60-80∙(1,77-0,6)+110 – 90∙(1,77-1,47) – 10∙(1,77-1,47)²/2=31,31 кНм

Задача 3
Задача 4

a = 6             m₁ = 6 кНм/м

b = 4,7          M₂ = 60 кН

c = 4             M₃ = 110 кН

d = 3

1)  0 ≤ x₁ ≤ d

M_x1 = -m₁∙x₁

x₁ = 0;             M_x1 = 0 кНм

x₁ = d;             M_x1 = -6∙3 = -18 кНм

2) d ≤ x₂ ≤ c + d

M_x2 = -m₁d + M₂ = -18 + 60 = 42 кНм

3) c + d ≤ x₃ ≤ b + c + d

M_x3 = - m₁d + M₂  + M₃ + m₁(x₃ – c – d) = -20 кН

x₃ = c + d;                   M_x3 = 42 + 110 = 152 кНм

x₃ = b + c + d;            M_x3 = 152 + 6 ∙4,7 = 182,2 кНм

4) b + c + d ≤  x₄ ≤ a + b + c + d

M_x4 = -m₁d + M₂ + M₃ + m₁b = 180,2 кНм

Задача 5