# Cut reaction

1st drawing: Beam with line load q and longitudinal force F as load (impressed values).
2nd drawing: Section reactions at an interface (normal force N, transverse force V, bending moment M)
Three line diagrams : The section reactions (internal forces) can be read off at any point x on the state lines .

The interface reactions are the cutting principle to be applied to the cut surfaces forces (so-called cutting forces ) and moments (so-called internal moments ). From these, the stresses in the imaginary cut surface (or in the undivided body at this point), which are the basis of strength and deformation studies , can be determined.

The internal forces in a solid rod-shaped component in a plane force system are:

If the rod-shaped component is in a spatial force system, the transverse force is represented with two components in an advantageous right-angled coordinate system (x-axis equals rod axis). The moment then has three components, two bending moments and a torsional moment rotating in the rod axis .

## Application of the cutting principle on a solid, rod-shaped component

The rod to be examined is mentally cut at any point. The two pattern pieces can be viewed individually. The cutting plane is usually flat and perpendicular to the rod axis. The influence of the cut-away part on the part to be examined is represented by the forces and moments applied at the interface - called sectional reactions or internal forces . The part to be examined remains in equilibrium due to the internal forces.

Alternatively, any part can be cut out by making two cuts. The cut sizes are then to be entered on both cut surfaces.

If you think of a gap between the two cut surfaces when cutting, then these can be viewed as two banks - a left and a right. The cut surface at the left end of a part is called the right or negative cut edge , and the right end is called the left or positive cut edge .

Straight rod with two interfaces.
The sectional reactions (forces and moments) shown here by arrows are shown as positive according to the usual sign convention.
The reference fiber is shown as a dashed line.

The intersection with the normal vector in the positive x-direction is the positive cutting edge . All internal forces point in a positive direction. The normal force N points in the positive x-direction, the transverse force Q in the positive z-direction and the bending moment has a positive direction of rotation (left or counterclockwise ) around the y-axis.

If the normal vector of the cutting surface points in the negative x-direction, one speaks accordingly of the negative cutting edge . In this case, all internal forces point in a negative direction, and the bending moment rotates around the y-axis in the negative direction of rotation (corresponding to clockwise ).

## Section reactions in the plane

The position of the local rod axis coordinate system is defined by the dashed fiber ( reference fiber ). It defines the direction and origin of the x, y and z axes as well as the position of the y and z axes. According to the usual convention : The z-axis points towards the dashed side. The x-axis is the member axis. The y-axis points out of the plane to the viewer. This results in a Cartesian right coordinate system .

• Shear Force - A force perpendicular to the part's x-axis. Common notations are V (vertical), F Q , Q (transverse), F y / F z (in y- / z-direction)
• Normal Force or Longitudinal Force - A force parallel to the part's x-axis. Notations are N, F N (normal = perpendicular to the cross section of the interface), F L (longitudinal)
• (Bending) moment - The moment acting at the interface . In the even case it is simply referred to as M.

All cutting reactions are vector quantities . This means that every force has a direction and every moment rotates either clockwise or counterclockwise.

A rod or a more complex object can also be cut free in a joint. The following applies:

• The reaction moment at a “moment joint” is zero.
• The lateral reaction force at a "lateral force joint" is equal to the force applied there.
• The normal reaction force at a "normal force joint" is equal to the force applied there.

## Cutting reactions in space

The lateral reaction force and the bending moment appear in the Cartesian coordinate system as two components each. The reaction moment contains a third component, namely one that rotates around the x-axis ( torsion moment , notations are M x , M T or T).

## State lines

If you move the imaginary interface along the component, the cutting reactions change. The sectional reactions shown in a diagram as a function of the interface location (coordinate x, see picture in the introduction ) are called state lines, since the sectional reactions correspond to the internal state ( mechanical stresses ) existing at this point . The points of greatest stress in a beam with a constant cross-section are clearly visible in the condition lines.

## Lines of influence

Construction of the #influence lines on which the bending moment (m) and the shear force (q) at point C of the single-span girder under a single load P that can be
shifted right-left can be read. The length c represents the value of P. The value of m and q is read off at the point below the force P marked accordingly in the diagram. However, this value applies to the internal forces at point C.

In contrast to the state line, an influence line does not represent a cutting reaction depending on the interface (coordinate x), but the dependence of a cutting reaction at a certain location (coordinate x 0 → the position of interface C in the figure ) on the variable location (coordinate x → in the figure represents the position of the force P ) of an external load.

A portable load are z. B. Subjects to bridges by a vehicle rolling over them. We are interested in the influence of the variable point of the load on the stress on the bridge at a certain point, which is expressed by the size of the cutting reactions at an imaginary cut there.

## Calculate the cutting reactions

There are several options for calculating cutting reactions. The equilibrium conditions must be observed for all of them. A general distinction is made between:

### Closed power corner

The force corner can be solved arithmetically or graphically.

The equilibrium conditions generally only provide unambiguous results for statically determined systems. In statically indeterminate systems, there are generally too many unknowns to apply without additional equations. In this case, e.g. B. Force or displacement method for the solution. The sectional reactions (on statically determined systems) are normally calculated using the equilibrium conditions. The equilibrium conditions for static systems state that

1. the sum of all forces is zero and${\ displaystyle \ sum {\ vec {F}} _ {i} = {\ vec {0}}}$
2. the sum of all moments (around any point) is zero .${\ displaystyle \ sum {\ vec {M}} _ {i} = {\ vec {0}}}$

With these equilibrium conditions, equations are created that make it possible to calculate the missing forces or moments. An intelligent selection of the sections (e.g. so that only one unknown quantity appears in an equation) can often reduce the computational effort, but this is not always possible.

The exact equations are given below.

#### Plane case (2D)

Here you can define 3 linearly independent moment equilibrium conditions, the easiest way is generally to define two points at infinity and thus you get two force component equilibrium conditions and one moment equilibrium condition:

${\ displaystyle \ Sigma F_ {ix} = 0}$
${\ displaystyle \ Sigma F_ {iz} = 0}$
${\ displaystyle \ Sigma M_ {i} ^ {(A)} = 0}$

Note: The index "A" in the moment equation indicates that the sum of the moments around a fictitious pivot point "A" is being considered here.

#### General case (3D)

Here z. B. three force component equilibrium conditions and three moment component equilibrium conditions:

${\ displaystyle \ Sigma F_ {ix} = 0}$
${\ displaystyle \ Sigma F_ {iy} = 0}$
${\ displaystyle \ Sigma F_ {iz} = 0}$
${\ displaystyle \ Sigma M_ {ix} ^ {(A1)} = 0}$
${\ displaystyle \ Sigma M_ {iy} ^ {(A2)} = 0}$
${\ displaystyle \ Sigma M_ {iz} ^ {(A3)} = 0}$

The points "A1", "A2", "A3" may also be identical, but also provide new linearly independent equations, since they consider the torques about a different axis. Alternatively, z. B. one also set up six moment equilibrium conditions, but here one must choose different points.

### Internal force differential equations

With this approach, differential equations that meet the equilibrium conditions are set up for the required internal forces and then solved with boundary conditions that match the system (for example: no torque transfer at a bearing at position x = 0).

In the shear-soft beam theory of the second order there are the following differential equations for the transverse components under Bernoulli's assumptions :

1. ${\ displaystyle {\ frac {\ mathrm {d} R (x)} {\ mathrm {d} x}} = - q (x)}$
2. ${\ displaystyle {\ frac {\ mathrm {d} M (x)} {\ mathrm {d} x}} = R (x) -N ^ {II} (x) \ cdot \ left [{\ frac {\ mathrm {d} w_ {v}} {\ mathrm {d} x}} + {\ frac {\ mathrm {d} w} {\ mathrm {d} x}} \ right] + m (x)}$
3. ${\ displaystyle {\ frac {\ mathrm {d} \ varphi (x)} {\ mathrm {d} x}} = - \ left [{\ frac {M (x)} {E \ cdot I (x)} } + \ kappa ^ {e} (x) \ right]}$
4. ${\ displaystyle {\ frac {\ mathrm {d} w (x)} {\ mathrm {d} x}} = \ varphi (x) + {\ frac {V (x)} {G {\ tilde {A} } (x)}}}$

With

• the running coordinate x along the beam axis
• the modulus of elasticity  E
• the shear modulus  G (term does not appear in the differential equations in rigid theory)
• the geometrical moment of inertia  I (x)
• R (x) is the transverse force (in first order theory , R (x) = V (x))
• V (x) is the shear force
• N II (x) the normal force according to the theory of the second order (in the theory of the first order this term does not appear in the differential equation)
• M (x) is the bending moment
• m (x) the insertion torque (bending load per unit length)
• φ (x) of the twist
• κ e (x) of the impressed curvature
• w (x) is the load due to deflection
• w v (x) deformation due to the deflection
• ${\ displaystyle {\ tilde {A}} (x)}$ the shear area (term does not appear in the rigid theory).

The first two differential equations follow solely from the equilibrium conditions, the first follows from the sum of the vertical forces is zero and the second from the knowledge that the sum of the moments must be zero. The last two differential equations are geometric differential relationships, the first derivative of the bending line is the inclination and the second derivative is the curvature, in addition two material equations are used here:

${\ displaystyle \ kappa (x) = {\ frac {M (x)} {E \ cdot I (x)}}}$and .${\ displaystyle {\ overline {\ gamma}} (x) = {\ frac {V (x)} {G {\ tilde {A}} (x)}}}$