How to calculate the concentrated force in the middle of hinge

1. One end fixed support, one end movable hinged support, the middle of which is added with concentrated load to calculate the support reaction:
RA = 5p / 16 RB = 11p / 16
fixed support is fixed, which limits the X, Y direction displacement and rotation in the plan. Hinged support is the limit line displacement, which can rotate. The setting of these two kinds of supports in the beam can adjust the reinforcement of the beam, The beam supported by fixed support can be regarded as continuous beam. There is negative bending moment at the support. Under uniform load, the bending moment in the middle of the span is smaller than that of the simply supported beam under the same condition. Therefore, the reinforcement at the bottom of the beam can be reced.

2. AC and ab are two force bars, and the restraint reaction of a and B points must be connected through AC and BC points
according to the force balance analysis, it can be seen that the restraint reaction direction of a point is from C to a, and the restraint reaction direction of B point is from B to C

PS: is this the basic examination question of injection electricity?

3.

This concentrated force acts on the middle hinge, not on the left and right arches, so the left and right arches are two force components. Similar to this (this is the stress of the right arch):

4.

The hinge problem is a very good physical problem

AB rod only receives the force acting on two points a and B respectively, with equal magnitude, opposite direction and along the rod. Why

Because AB rod only contacts with the outside at a and B points, it can only bear force at a and B points

The two forces must be equal in magnitude and opposite in direction, otherwise AB will move

These two forces must be along the direction of the rod, otherwise AB will rotate

BC is the same

why does the force direction of rod BC to B point to B

because point B is subjected to three forces, the directions of AB rod, BC rod and rope are in AB rod direction, BC rod direction and vertical downward direction respectively

First of all, assume that the force of bar AB on point B is pointing to B along the bar, that is, diagonally downward, as shown in figure (1). Then combine this force with the tension of the rope, the resultant force is f, then the force of BC rod on B must be equal to F, and the direction is opposite. And you will find that the opposite direction of F is above the AB bar (which means that point C is above point a, which is contradictory to the figure in the question)

Therefore, the force of AB rod on B must be along AB and point to a. So the stress is shown in figure (2)

5. When the action is a concentrated force, it is regarded as acting on the basic part. When the action is a couple, it is divided into two cases: acting on the left and right of the node

6. Unknown_Error

7. A. When CD is in tension, the elongation along CD is D, and the component in AB vertical direction is a

8. Force method or displacement method. Did you forget. At the hinge, the deformation is coordinated. It can be solved by establishing the equation.

9.

Solution: let the reaction force at the support of simply supported beam be r, the uniform load on the beam be q, and the calculated span length of the beam be L

According to the principle of static equilibrium, it is concluded that: take the length of the calculation section of the beam as X, take the detached body, and set the anti clockwise bending moment as positive, take the moment on the X plane of the calculation point, and the resultant bending moment is zero

have MX = rx-qx ^ 2 / 2 = (qlx / 2) - (QX ^ 2 / 2)

take the derivative of X, have the first derivative M '= QL / 2-qx

have the second derivative M' = - Q < 0, so it can be determined that M has a maximum value

Let the first derivative be equal to zero, there is QL / 2-qx = 0

so, x = L / 2 brings it back to MX, there is

mmax = m (x = L / 2) = (QL ^ 2 / 4) - (QL ^ 2 / 8) = QL ^ 2 / 8

so the middle bending moment of simply supported beam under uniform load is 1 / 8ql2

extended data:

bending moment diagram features

there are two key points in the drawing of bending moment diagram: one is to accurately draw the shape of the curve, that is, to determine the graphic features of the bending moment diagram; the other is to determine the position of the curve, that is, to determine the position of the plane curve after the shape and size of the curve are known, This requires that the position of any two points on the curve should be determined first. The position of the two points here refers to the bending moment value of a certain two sections

It can be seen that the drawing of bending moment diagram mainly refers to the following two tasks:

(1) determine the graphic features and eigenvalues

(2) the bending moment values of two sections are obtained

Be familiar with the bending moment diagram characteristics of single span beam under various independent loads: for example, the bending moment diagram of cantilever beam under a concentrated load is characterized by a right triangle; The bending moment diagram of a cantilever beam is a curved triangle when the uniform load acts on the whole length. Bending moment diagram of single span beam under one kind of load

2. Determination of bending moment value at both ends of a certain section of a member there are generally three situations as follows:

(1) non hinged beam section: generally, the bending moment value at both ends of the beam section should be calculated first

(2) there is a hinge in the middle of the beam section: since the bending moment at the hinge without external force moment is known to be zero, it is only necessary to calculate the bending moment of another section

(3) there are two hinges in the middle of the beam segment: the bending moment at the two hinges is zero, so the bending moment diagram can be drawn directly according to the characteristics of the bending moment diagram of the simply supported beam

10. Whatever you want, you can count it on the left or right. For example, if you think it's on the left, then it doesn't count on the right