Quadratic equations.
Quadratic equation- algebraic equation of general form
where x is a free variable,
a, b, c, are coefficients, and
Expression 
called a square trinomial.
Methods for solving quadratic equations.
1. METHOD : Factoring the left side of the equation.
Let's solve the equation x 2 + 10x - 24 = 0. Let's factorize the left side:
x 2 + 10x - 24 = x 2 + 12x - 2x - 24 = x(x + 12) - 2(x + 12) = (x + 12)(x - 2).
Therefore, the equation can be rewritten as follows:
(x + 12)(x - 2) = 0
Since the product is zero, then at least one of its factors is zero. Therefore, the left side of the equation becomes zero at x = 2, and also when x = - 12. This means that the number 2 And - 12 are the roots of the equation x 2 + 10x - 24 = 0.
2. METHOD : Method for selecting a complete square.
Let's solve the equation x 2 + 6x - 7 = 0. Select a complete square on the left side.
To do this, we write the expression x 2 + 6x in the following form:
x 2 + 6x = x 2 + 2 x 3.
In the resulting expression, the first term is the square of the number x, and the second is the double product of x by 3. Therefore, to get a complete square, you need to add 3 2, since
x 2 + 2 x 3 + 3 2 = (x + 3) 2.
Let us now transform the left side of the equation
x 2 + 6x - 7 = 0,
adding to it and subtracting 3 2. We have:
x 2 + 6x - 7 = x 2 + 2 x 3 + 3 2 - 3 2 - 7 = (x + 3) 2 - 9 - 7 = (x + 3) 2 - 16.
Thus, this equation can be written as follows:
(x + 3) 2 - 16 =0, (x + 3) 2 = 16.
Hence, x + 3 - 4 = 0, x 1 = 1, or x + 3 = -4, x 2 = -7.
3. METHOD :Solving quadratic equations using the formula.
Let's multiply both sides of the equation
ax 2 + bx + c = 0, a ≠ 0
on 4a and sequentially we have:
4a 2 x 2 + 4abx + 4ac = 0,
((2ax) 2 + 2ax b + b 2) - b 2 + 4ac = 0,
(2ax + b) 2 = b 2 - 4ac,
2ax + b = ± √ b 2 - 4ac,
2ax = - b ± √ b 2 - 4ac,
Examples.
A) Let's solve the equation: 4x 2 + 7x + 3 = 0.
a = 4, b = 7, c = 3, D = b 2 - 4ac = 7 2 - 4 4 3 = 49 - 48 = 1,
D > 0, two different roots;
Thus, in the case of a positive discriminant, i.e. at
b 2 - 4ac >0, the equation ax 2 + bx + c = 0 has two different roots.
b) Let's solve the equation: 4x 2 - 4x + 1 = 0,
a = 4, b = - 4, c = 1, D = b 2 - 4ac = (-4) 2 - 4 4 1= 16 - 16 = 0,
D = 0, one root;
So, if the discriminant is zero, i.e. b 2 - 4ac = 0, then the equation
ax 2 + bx + c = 0 has a single root
V) Let's solve the equation: 2x 2 + 3x + 4 = 0,
a = 2, b = 3, c = 4, D = b 2 - 4ac = 3 2 - 4 2 4 = 9 - 32 = - 13, D< 0.
This equation has no roots.
So, if the discriminant is negative, i.e. b 2 - 4ac< 0 , the equation
ax 2 + bx + c = 0 has no roots.
Formula (1) of the roots of a quadratic equation ax 2 + bx + c = 0 allows you to find roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed verbally as follows: the roots of a quadratic equation are equal to a fraction whose numerator is equal to the second coefficient taken with the opposite sign, plus minus the square root of the square of this coefficient without quadruple the product of the first coefficient by the free term, and the denominator is double the first coefficient.
4. METHOD: Solving equations using Vieta's theorem.
As is known, the reduced quadratic equation has the form
x 2 + px + c = 0.(1)
Its roots satisfy Vieta’s theorem, which, when a =1 looks like
x 1 x 2 = q,
x 1 + x 2 = - p
From this we can draw the following conclusions (from the coefficients p and q we can predict the signs of the roots).
a) If the half-member q given equation (1) is positive ( q > 0), then the equation has two roots of equal sign and this depends on the second coefficient p. If R< 0 , then both roots are negative if R< 0 , then both roots are positive.
For example,
x 2 – 3x + 2 = 0; x 1 = 2 And x 2 = 1, because q = 2 > 0 And p = - 3< 0;
x 2 + 8x + 7 = 0; x 1 = - 7 And x 2 = - 1, because q = 7 > 0 And p= 8 > 0.
b) If a free member q given equation (1) is negative ( q< 0 ), then the equation has two roots of different sign, and the larger root will be positive if p< 0 , or negative if p > 0 .
For example,
x 2 + 4x – 5 = 0; x 1 = - 5 And x 2 = 1, because q= - 5< 0 And p = 4 > 0;
x 2 – 8x – 9 = 0; x 1 = 9 And x 2 = - 1, because q = - 9< 0 And p = - 8< 0.
Examples.
1) Let's solve the equation 345x 2 – 137x – 208 = 0.
Solution. Because a + b + c = 0 (345 – 137 – 208 = 0), That
x 1 = 1, x 2 = c/a = -208/345.
Answer: 1; -208/345.
2) Solve the equation 132x 2 – 247x + 115 = 0.
Solution. Because a + b + c = 0 (132 – 247 + 115 = 0), That
x 1 = 1, x 2 = c/a = 115/132.
Answer: 1; 115/132.
B. If the second coefficient b = 2k is an even number, then the root formula
Example.
Let's solve the equation 3x2 - 14x + 16 = 0.
Solution. We have: a = 3, b = - 14, c = 16, k = - 7;
D = k 2 – ac = (- 7) 2 – 3 16 = 49 – 48 = 1, D > 0, two different roots;
Answer: 2; 8/3
IN. Reduced equation
x 2 + px + q= 0
coincides with a general equation in which a = 1, b = p And c = q. Therefore, for the reduced quadratic equation, the root formula is
Takes the form:
Formula (3) is especially convenient to use when R- even number.
Example. Let's solve the equation x 2 – 14x – 15 = 0.
Solution. We have: x 1.2 =7±
Answer: x 1 = 15; x 2 = -1.
5. METHOD: Solving equations graphically.
Example. Solve the equation x2 - 2x - 3 = 0.
Let's plot the function y = x2 - 2x - 3
1) We have: a = 1, b = -2, x0 = = 1, y0 = f(1) = 12 - 2 - 3 = -4. This means that the vertex of the parabola is the point (1; -4), and the axis of the parabola is the straight line x = 1.
2) Take two points on the x-axis that are symmetrical about the axis of the parabola, for example, points x = -1 and x = 3.
We have f(-1) = f(3) = 0. Let’s construct points (-1; 0) and (3; 0) on the coordinate plane.
3) Through the points (-1; 0), (1; -4), (3; 0) we draw a parabola (Fig. 68).
The roots of the equation x2 - 2x - 3 = 0 are the abscissas of the points of intersection of the parabola with the x-axis; This means that the roots of the equation are: x1 = - 1, x2 - 3.
Let us recall the basic properties of degrees. Let a > 0, b > 0, n, m be any real numbers. Then
1) a n a m = a n+m
2) \(\frac(a^n)(a^m) = a^(n-m) \)
3) (a n) m = a nm
4) (ab) n = a n b n
5) \(\left(\frac(a)(b) \right)^n = \frac(a^n)(b^n) \)
7) a n > 1, if a > 1, n > 0
8) a n 1, n
9) a n > a m if 0
In practice, functions of the form y = a x are often used, where a is a given positive number, x is a variable. Such functions are called indicative. This name is explained by the fact that the argument of the exponential function is the exponent, and the base of the exponent is the given number.
Definition. An exponential function is a function of the form y = a x, where a is a given number, a > 0, \(a \neq 1\)
The exponential function has the following properties
1) The domain of definition of the exponential function is the set of all real numbers.
This property follows from the fact that the power a x where a > 0 is defined for all real numbers x.
2) The set of values of the exponential function is the set of all positive numbers.
To verify this, you need to show that the equation a x = b, where a > 0, \(a \neq 1\), has no roots if \(b \leq 0\), and has a root for any b > 0 .
3) The exponential function y = a x is increasing on the set of all real numbers if a > 1, and decreasing if 0. This follows from the properties of degree (8) and (9)
Let's construct graphs of exponential functions y = a x for a > 0 and for 0. Using the considered properties, we note that the graph of the function y = a x for a > 0 passes through the point (0; 1) and is located above the Ox axis.
If x 0.
If x > 0 and |x| increases, the graph quickly rises.
Graph of the function y = a x at 0 If x > 0 and increases, then the graph quickly approaches the Ox axis (without crossing it). Thus, the Ox axis is the horizontal asymptote of the graph.
If x 
Exponential equations
Let's consider several examples of exponential equations, i.e. equations in which the unknown is contained in the exponent. Solving exponential equations often comes down to solving the equation a x = a b where a > 0, \(a \neq 1\), x is an unknown. This equation is solved using the power property: powers with the same base a > 0, \(a \neq 1\) are equal if and only if their exponents are equal.
Solve equation 2 3x 3 x = 576
Since 2 3x = (2 3) x = 8 x, 576 = 24 2, the equation can be written as 8 x 3 x = 24 2, or as 24 x = 24 2, from which x = 2.
Answer x = 2
Solve the equation 3 x + 1 - 2 3 x - 2 = 25
Taking the common factor 3 x - 2 out of brackets on the left side, we get 3 x - 2 (3 3 - 2) = 25, 3 x - 2 25 = 25,
whence 3 x - 2 = 1, x - 2 = 0, x = 2
Answer x = 2
Solve the equation 3 x = 7 x
Since \(7^x \neq 0 \) , the equation can be written in the form \(\frac(3^x)(7^x) = 1 \), from which \(\left(\frac(3)( 7) \right) ^x = 1 \), x = 0
Answer x = 0
Solve the equation 9 x - 4 3 x - 45 = 0
By replacing 3 x = t, this equation is reduced to the quadratic equation t 2 - 4t - 45 = 0. Solving this equation, we find its roots: t 1 = 9, t 2 = -5, whence 3 x = 9, 3 x = -5 .
The equation 3 x = 9 has a root x = 2, and the equation 3 x = -5 has no roots, since the exponential function cannot take negative values.
Answer x = 2
Solve equation 3 2 x + 1 + 2 5 x - 2 = 5 x + 2 x - 2
Let's write the equation in the form
3 2 x + 1 - 2 x - 2 = 5 x - 2 5 x - 2, whence
2 x - 2 (3 2 3 - 1) = 5 x - 2 (5 2 - 2)
2 x - 2 23 = 5 x - 2 23
\(\left(\frac(2)(5) \right) ^(x-2) = 1 \)
x - 2 = 0
Answer x = 2
Solve equation 3 |x - 1| = 3 |x + 3|
Since 3 > 0, \(3 \neq 1\), then the original equation is equivalent to the equation |x-1| = |x+3|
By squaring this equation, we obtain its corollary (x - 1) 2 = (x + 3) 2, from which
x 2 - 2x + 1 = x 2 + 6x + 9, 8x = -8, x = -1
Checking shows that x = -1 is the root of the original equation.
Answer x = -1
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Certificates and guarantees
I. ax 2 =0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.
Solve equations.
2x·(x+3)=6x-x 2 .
Solution. Let's open the brackets by multiplying 2x for each term in brackets:
2x 2 +6x=6x-x 2 ; We move the terms from the right side to the left:
2x 2 +6x-6x+x 2 =0; Here are similar terms:
3x 2 =0, hence x=0.
Answer: 0.
II. ax 2 +bx=0 –incomplete quadratic equation (c=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.
5x 2 -26x=0.
Solution. Let's take out the common factor X outside the brackets:
x(5x-26)=0; each factor can be equal to zero:
x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x=5.2.
Answer: 0; 5,2.
Example 3. 64x+4x 2 =0.
Solution. Let's take out the common factor 4x outside the brackets:
4x(16+x)=0. We have three factors, 4≠0, therefore, or x=0 or 16+x=0. From the last equality we get x=-16.
Answer: -16; 0.
Example 4.(x-3) 2 +5x=9.
Solution. Applying the formula for the square of the difference of two expressions, we will open the brackets:
x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Let us present similar terms:
x 2 -x=0; we'll take it out X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.
Answer: 0; 1.
III. ax 2 +c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 =-c → x 2 =-c/a.
If (-c/a)<0 , then there are no real roots. If (-с/а)>0
Example 5. x 2 -49=0.
Solution.
x 2 =49, from here x=±7. Answer:-7; 7.
Example 6. 9x 2 -4=0.
Solution.
Often you need to find the sum of squares (x 1 2 +x 2 2) or the sum of cubes (x 1 3 +x 2 3) of the roots of a quadratic equation, less often - the sum of the reciprocal values of the squares of the roots or the sum of arithmetic square roots of the roots of a quadratic equation:
Vieta's theorem can help with this:
x 2 +px+q=0
x 1 + x 2 = -p; x 1 ∙x 2 =q.
Let's express through p And q:
1) sum of squares of the roots of the equation x 2 +px+q=0;
2) sum of cubes of the roots of the equation x 2 +px+q=0.
Solution.
1) Expression x 1 2 +x 2 2 obtained by squaring both sides of the equation x 1 + x 2 = -p;
(x 1 +x 2) 2 =(-p) 2 ; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2 ; we express the required amount: x 1 2 +x 2 2 =p 2 -2x 1 x 2 =p 2 -2q. We got a useful equality: x 1 2 +x 2 2 =p 2 -2q.
2) Expression x 1 3 +x 2 3 Let us represent the sum of cubes using the formula:
(x 1 3 +x 2 3)=(x 1 +x 2)(x 1 2 -x 1 x 2 +x 2 2)=-p·(p 2 -2q-q)=-p·(p 2 -3q).
Another useful equation: x 1 3 +x 2 3 = -p·(p 2 -3q).
Examples.
3) x 2 -3x-4=0. Without solving the equation, calculate the value of the expression x 1 2 +x 2 2.
Solution.
x 1 +x 2 =-p=3, and the work x 1 ∙x 2 =q=in example 1) equality:
x 1 2 +x 2 2 =p 2 -2q. We have -p=x 1 +x 2 = 3 → p 2 =3 2 =9; q= x 1 x 2 = -4. Then x 1 2 +x 2 2 =9-2·(-4)=9+8=17.
Answer: x 1 2 +x 2 2 =17.
4) x 2 -2x-4=0. Calculate: x 1 3 +x 2 3 .
Solution.
By Vieta's theorem, the sum of the roots of this reduced quadratic equation is x 1 +x 2 =-p=2, and the work x 1 ∙x 2 =q=-4. Let's apply what we received ( in example 2) equality: x 1 3 +x 2 3 =-p·(p 2 -3q)= 2·(2 2 -3·(-4))=2·(4+12)=2·16=32.
Answer: x 1 3 +x 2 3 =32.
Question: what if we are given an unreduced quadratic equation? Answer: it can always be “reduced” by dividing term by term by the first coefficient.
5) 2x 2 -5x-7=0. Without deciding, calculate: x 1 2 +x 2 2.
Solution. We are given a complete quadratic equation. Divide both sides of the equality by 2 (the first coefficient) and obtain the following quadratic equation: x 2 -2.5x-3.5=0.
According to Vieta's theorem, the sum of the roots is equal to 2,5 ; the product of the roots is equal -3,5 .
We solve it in the same way as the example 3) using the equality: x 1 2 +x 2 2 =p 2 -2q.
x 1 2 +x 2 2 =p 2 -2q= 2,5 2 -2∙(-3,5)=6,25+7=13,25.
Answer: x 1 2 + x 2 2 = 13,25.
6) x 2 -5x-2=0. Find:
Let us transform this equality and, using Vieta’s theorem, replace the sum of roots through -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 =p 2 -2q.
In our example x 1 +x 2 =-p=5; x 1 ∙x 2 =q=-2. We substitute these values into the resulting formula:
7) x 2 -13x+36=0. Find:
Let's transform this sum and get a formula that can be used to find the sum of arithmetic square roots from the roots of a quadratic equation.
We have x 1 +x 2 =-p=13; x 1 ∙x 2 =q=36. We substitute these values into the resulting formula:
Advice : always check the possibility of finding the roots of a quadratic equation using a suitable method, because 4 reviewed useful formulas allow you to quickly complete a task, especially in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example we select the roots using Vieta’s theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Certainly, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!
I. Vieta's theorem for the reduced quadratic equation.
Sum of roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 = -p; x 1 ∙x 2 =q.
Find the roots of the given quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), second coefficient p=-1, and the free member q=-30. First, let's make sure that this equation has roots, and that the roots (if any) will be expressed in integers. To do this, it is enough that the discriminant be a perfect square of an integer.
Finding the discriminant D=b 2 — 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to Vieta’s theorem, the sum of the roots must be equal to the second coefficient taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 +x 2 =1; x 1 ∙x 2 =-30. We need to choose two numbers such that their product is equal to -30 , and the amount is unit. These are numbers -5 And 6 . Answer: -5; 6.
Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Let's make sure that there are integer roots. Let's find the discriminant D 1 D 1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , which means that the roots of this equation are integers. Let us select the roots using Vieta’s theorem: the sum of the roots is equal to –р=-6, and the product of the roots is equal to q=8. These are numbers -4 And -2 .
In fact: -4-2=-6=-р; -4∙(-2)=8=q. Answer: -4; -2.
Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free member q=-4. Let's find the discriminant D 1, since the second coefficient is an even number. D 1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of the number, so we do conclusion: The roots of this equation are not integers and cannot be found using Vieta’s theorem. This means that we solve this equation, as usual, using formulas (in this case, using formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 =-7, x 2 =4.
Solution. The required equation will be written in the form: x 2 +px+q=0, and, based on Vieta’s theorem –p=x 1 +x 2=-7+4=-3 → p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x 2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for a complete quadratic equation ax 2 +bx+c=0.
The sum of the roots is minus b, divided by A, the product of the roots is equal to With, divided by A:
x 1 + x 2 = -b/a; x 1 ∙x 2 =c/a.
Example 6). Find the sum of the roots of a quadratic equation 2x 2 -7x-11=0.
Solution.
We make sure that this equation will have roots. To do this, it is enough to create an expression for the discriminant, and, without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . Now let's use theorem Vieta for complete quadratic equations.
x 1 +x 2 =-b:a=- (-7):2=3,5.
Example 7). Find the product of the roots of a quadratic equation 3x 2 +8x-21=0.
Solution.
Let's find the discriminant D 1, since the second coefficient ( 8 ) is an even number. D 1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to Vieta’s theorem, the product of roots x 1 ∙x 2 =c:a=-21:3=-7.
I. ax 2 +bx+c=0– general quadratic equation
Discriminant D=b 2 - 4ac.
If D>0, then we have two real roots:
If D=0, then we have a single root (or two equal roots) x=-b/(2a).
If D<0, то действительных корней нет.
Example 1) 2x 2 +5x-3=0.
Solution. a=2; b=5; c=-3.
D=b 2 - 4ac=5 2 -4∙2∙(-3)=25+24=49=7 2 >0; 2 real roots.
4x 2 +21x+5=0.
Solution. a=4; b=21; c=5.
D=b 2 - 4ac=21 2 - 4∙4∙5=441-80=361=19 2 >0; 2 real roots.
II. ax 2 +bx+c=0 – quadratic equation of particular form with even second
coefficient b
Example 3) 3x 2 -10x+3=0.
Solution. a=3; b=-10 (even number); c=3.
Example 4) 5x 2 -14x-3=0.
Solution. a=5; b= -14 (even number); c=-3.
Example 5) 71x 2 +144x+4=0.
Solution. a=71; b=144 (even number); c=4.
Example 6) 9x 2 -30x+25=0.
Solution. a=9; b=-30 (even number); c=25.
III. ax 2 +bx+c=0 – quadratic equation private type provided: a-b+c=0.
The first root is always equal to minus one, and the second root is always equal to minus With, divided by A:
x 1 =-1, x 2 =-c/a.
Example 7) 2x 2 +9x+7=0.
Solution. a=2; b=9; c=7. Let's check the equality: a-b+c=0. We get: 2-9+7=0 .
Then x 1 =-1, x 2 =-c/a=-7/2=-3.5. Answer: -1; -3,5.
IV. ax 2 +bx+c=0 – quadratic equation of a particular form subject to : a+b+c=0.
The first root is always equal to one, and the second root is equal to With, divided by A:
x 1 =1, x 2 =c/a.
Example 8) 2x 2 -9x+7=0.
Solution. a=2; b=-9; c=7. Let's check the equality: a+b+c=0. We get: 2-9+7=0 .
Then x 1 =1, x 2 =c/a=7/2=3.5. Answer: 1; 3,5.
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The foundations of power transmission line supports for energy construction face a responsible task - to maintain the stability and strength of power transmission line supports for many years in different climatic conditions, at any time of the year and in any weather. Therefore, very high demands are placed on support foundations. Before shipping to the customer, the foundations of FP2.7x2.7-A supports for metal supports of 220 kV single-circuit overhead lines, 330 kV single-circuit overhead lines are tested according to various parameters, for example, the degree of stability, strength, durability and wear resistance, resistance to negative temperatures and atmospheric influences . Before welding, joint parts must be free of rust. Reinforced concrete foundations with a concrete protective layer thickness of less than 30 mm, as well as foundations installed in aggressive soils, must be protected by waterproofing.
During operation, foundations FP2.7x2.7-A for metal supports of 220 kV single-circuit overhead lines, 330 kV single-circuit overhead lines are subject to careful supervision, especially in the first years of operation of the overhead line. One of the most serious defects in the construction of foundations, difficult to eliminate under operating conditions, is a violation of technological standards during their manufacture: the use of low-quality or poorly washed gravel, violation of proportions when preparing a concrete mixture, etc. An equally serious defect is layered concreting of foundations, when individual elements of the same foundation are concreted at different times without prior surface preparation. In this case, the concrete of one foundation element does not set with another, and destruction of the foundation may occur under external loads that are significantly less than the calculated ones.
When making reinforced concrete foundations for supports, standards are also sometimes violated: low-quality concrete is used, reinforcement is laid in the wrong sizes as provided for by the project. During the construction of power lines on prefabricated or piled reinforced concrete foundations, serious defects may occur that are not allowed by energy construction. Such defects include the installation of broken reinforced concrete foundations, their insufficient penetration into the ground (especially when installing supports on the slopes of hills and ravines), inappropriate compaction during backfilling, installation of prefabricated foundations of smaller sizes, etc. Installation defects include incorrect installation of reinforced concrete foundations, in which individual prefabricated foundations intended as the base of a metal support have different vertical elevations or shifts of individual foundations in plan. If improperly unloaded, the foundations FP2.7x2.7-A for metal supports of 220 kV single-circuit overhead lines, 330 kV single-circuit overhead lines can be damaged, concrete chipping and reinforcement may be exposed. During the acceptance process, special attention should be paid to the compliance of the anchor bolts and their nuts with the design dimensions.
Under operating conditions, reinforced concrete foundations FP2.7x2.7-A for metal supports of 220 kV single-circuit overhead lines, 330 kV single-circuit overhead lines are damaged both from environmental influences and from large external loads. The reinforcement of foundations with a porous concrete structure is damaged by the aggressive effects of groundwater. Cracks that form on the surface of foundations, when exposed to operational alternating loads, as well as wind, moisture and low temperature, expand, which ultimately leads to the destruction of concrete and exposure of reinforcement. In areas located near chemical plants, anchor bolts and the upper part of metal footrests quickly deteriorate.
Breakage of the support foundation can also occur as a result of its misalignment with the racks, which causes large bending moments. A similar breakdown can occur when the base of the foundation is washed away by groundwater and deviates from its vertical position.
During the acceptance process, FP2.7x2.7-A foundations for metal supports of 220 kV single-circuit overhead lines, 330 kV single-circuit overhead lines are checked for compliance with the design, laying depth, quality of concrete, quality of welding of working reinforcement and anchor bolts, presence and quality of protection against the action of aggressive waters . The vertical marks of the foundations are measured and the location of the anchor bolts is checked according to the template. If any non-compliance with the standards is detected, all defects are eliminated before backfilling the pits. Foundations that have chipped concrete and exposed reinforcement in the upper part are repaired. To do this, a concrete frame 10-20 cm thick is installed, buried 20-30 cm below ground level. It should be borne in mind that energy construction does not allow a frame made of slag concrete, since the slag contains an admixture of sulfur, which causes intense corrosion of reinforcement and anchors. bolts In case of more significant damage to foundations (including monolithic ones), the damaged part is covered with reinforcement welded to the reinforcement of the main foundation, and after installation of the formwork it is concreted.
