An example of the calculation of a statically uncertain system in torsion. Solving typical problems on strength of materials
Torsion of a rod of circular cross-section – problem condition
Four external torsional moments are applied to a steel shaft of constant cross-section (Fig. 3.8): kN m; kN m; kN m; kNm. Lengths of rod sections: m; m, m, m. Required: construct a diagram of torques, determine the diameter of the shaft at kN/cm2 and construct a diagram of the twist angles of the cross sections of the rod.
Torsion of a round rod - design diagram
Rice. 3.8
Solution to the problem of torsion of a round rod
Determine the reactive torque arising in a rigid seal
Let's designate the moment in the embedding and direct it, for example, counterclockwise (when looking towards the z-axis).
Let's write down the balance equation for the shaft. In this case, we will use the following sign rule: external twisting moments (active moments, as well as the reactive moment in the seal), rotating the shaft counterclockwise (when looking at it towards the z axis), are considered positive.
The plus sign in the expression we obtained indicates that we guessed the direction of the reactive torque arising in the seal.
We build a diagram of torques
Let us recall that the internal torque arising in a certain cross section of the rod is equal to the algebraic sum of the external twisting moments applied to any of the parts of the rod under consideration (that is, acting to the left or to the right of the section made). In this case, the external twisting moment, which rotates the part of the rod under consideration counterclockwise (when looking at the cross section), is included in this algebraic sum with a “plus” sign, and along the way – with a “minus” sign.
Accordingly, the positive internal torque counteracting the external twisting moments is directed clockwise (when looking at the cross section), and the negative one is counterclockwise.
We divide the length of the rod into four sections (Fig. 3.8, a). The boundaries of the sections are those sections in which external moments are applied.
We make one section at a random location in each of the four sections of the rod.
Section 1 – 1. Let’s mentally discard (or cover with a piece of paper) the left side of the rod. To balance the twisting moment kN m, an equal and oppositely directed torque must arise in the cross section of the rod. Taking into account the above mentioned sign rule
kNm.
Sections 2 – 2 and 3 – 3:
Section 4 – 4. To determine the torque, in section 4 – 4 we discard the right side of the rod. Then
kNm.
It is easy to verify that the result obtained will not change if we discard now not the right, but the left part of the rod. We get
To construct a diagram of torques, draw a thin line along the axis parallel to the rod axis z (Fig. 3.8, b). The calculated values of torques on the selected scale and taking into account their sign are plotted from this axis. Within each section of the rod, the torque is constant, so we seem to “shade” the corresponding section with vertical lines. Let us recall that each segment of the “hatching” (the ordinate of the diagram) gives, on the accepted scale, the value of the torque in the corresponding cross section of the rod. We outline the resulting diagram with a thick line.
Note that in the places where external twisting moments are applied on the diagram, we received an abrupt change in the internal torque by the value of the corresponding external torque.
Determine the shaft diameter from the strength condition
The torsional strength condition has the form
,
Where 
– polar moment of resistance (moment of resistance during torsion).
The greatest absolute value of torque occurs in the second section of the shaft: 
kN cm
Then the required shaft diameter is determined by the formula
cm.
Rounding the resulting value to the standard value, we take the shaft diameter to be equal to mm.
We determine the angles of twist of cross sections A, B, C, D and E and construct a diagram of the angles of twist
First, we calculate the torsional stiffness of the rod, where G is the shear modulus, and 
– polar moment of inertia. We get
The angles of twist in individual sections of the rod are equal:
glad;
glad;
glad;
glad.
The angle of twist in the embedment is zero, that is. Then
The diagram of twist angles is shown in Fig. 3.8, c. Note that within the length of each section of the shaft, the twist angle changes according to a linear law.
An example of a problem on torsion of a “round” rod for independent solution
Conditions for the problem of torsion of a “round” rod
A steel rod (shear modulus kN/cm2) of circular cross-section, rigidly clamped at one end, is twisted by four moments (Fig. 3.7).
Required:
· construct a diagram of torques;
· at a given permissible shear stress kN/cm2, from the strength condition, determine the diameter of the shaft, rounding it to the nearest of the following values 30, 35, 40, 45, 50, 60, 70, 80, 90, 100, 200 mm;
· construct a diagram of the angles of twist of the cross sections of the rod.
Variants of calculation schemes for the problem of torsion of a round rod for independent solution
An example of a problem on torsion of a round rod - initial conditions for independent solution
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When calculating the torsion of straight bars rigidly clamped at one end, as well as when calculating shafts (which are rotating bars loaded with mutually balanced torsional moments), the values of torques in cross sections can be determined using equilibrium equations alone (by the method of sections). Consequently, such problems are statically definable.
Torsional design problems are statically indeterminable if the torques occurring in the cross sections of the twisted bars cannot be determined using equilibrium equations alone. To solve these problems, in addition to the equilibrium equations compiled for the system as a whole or its cut-off part, it is also necessary to compile displacement equations based on consideration of the nature of the deformation of the system.
Let us consider, as an example, a beam of circular cross-section, rigidly embedded at both ends and loaded with a moment ZL at a distance a from the left end (Fig. 23.6, a).
To solve this problem, you can create only one equilibrium equation - in the form of the sum of moments about the axis of the beam being equal to zero:
where and are the reactive torsional moments arising in the seals.
An additional equation for solving the problem under consideration can be obtained as follows. Let's discard the left support fastening of the beam, but leave the right one (Fig. 23.6, b).
The rotation of the left end of the beam obtained in this way should be equal to zero, i.e., since in reality this end is rigidly fixed and cannot be rotated.
Based on the principle of independence of the action of forces, the displacement equation has the form
Here is the angle of rotation of the left end of the beam due to the action of an external twisting moment (Fig. 23.6, c); - angle of rotation of the left end due to the action of an external moment (Fig. 23.6, d).
Using the second of formulas (14.6), taking into account that the right end of the beam does not rotate (i.e.), and using formula (13.6) we find
Let's substitute these values into the displacement equation:
From the equilibrium equation
After determining the moments, the torque diagram can be constructed in the usual way, i.e., as for a statically determinate beam (Fig. 23.6, e). For the problem considered, this diagram is presented in Fig. 23.6, e.
A visual representation of the change in the angles of rotation of the cross sections of a beam along its length is given by the diagram of the angles of rotation (sometimes called the diagram of the angles of twist). Each ordinate of this diagram gives on the accepted scale the value of the angle of rotation of the corresponding cross-section of the beam.
Let's construct such a diagram for a beam according to Fig. 23.6, d, taking into account that the value has already been found and the torque diagram has been constructed (see Fig. 23.6, f). The rightmost section A of the beam is motionless, i.e., an arbitrary cross section belonging to section AC and spaced at a distance from the right end will rotate by an angle [see. second of formulas (14.6)]
Here is the angle of twist in a section of length determined by formula (13.6).
Thus, the angles of rotation change according to a linear law depending on the distance. Substituting into the resulting expression, we find the angle of rotation of section C:
Note that always when a beam of constant cross-section is loaded with concentrated torsional moments, the diagram of the angles of rotation of the cross sections on each section of the beam is linear.
To construct a diagram in the section NE, we calculate the angle of rotation of section B. Based on the second of formulas (14.6) and formula (13.6)
This result confirms the correctness of the solution to the problem, since according to the condition, section B is rigidly sealed. Thus, in addition to purely illustrative value, constructing a diagram of the rotation angles of cross sections can be considered as a method of monitoring the solution of some statically indeterminate problems.
The diagram of rotation angles constructed from the obtained values is shown in Fig. 23.6, w.
When several external torsional moments are applied to a beam, as well as for beams that have different cross sections in certain sections, an additional equation is drawn up in a manner similar to that shown (see example 5.6).
When calculating cylindrical springs, along with statically determinate problems, there are also statically indeterminate problems.
If the ends of the spring are not fixed and can move freely along the axis of the spring, or if only one end is fixed, then the problem of calculating such a spring is statically determinate. If both ends of the spring are fixedly fixed, then the problem of its calculation is statically indeterminate. To solve it, it is necessary to create an additional displacement equation. The compilation of this equation is similar to the compilation of the equation used in solving problems of calculating a straight rod fixed at both ends for external loads acting along its axis. The composition of additional equations for this type of problem is discussed above in § 9.2 (see also example 3.6).
4.4. Statically indeterminate torsion problems
Such problems usually arise if the movement of the shaft is limited in some sections, for example, (Fig. 4.9), when its ends are pinched. IN
one equilibrium equation: : 
there are two unknown moments in the supports, so the problem is statically indeterminate. To solve it, we create an additional displacement equation. Let's consider the displacements (rotation angles) of the sections that are the boundaries of the shaft sections..gif" width="99" height="27 src=">.
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Since the shaft section is pinched, then from: https://pandia.ru/text/78/579/images/image011_42.gif" align="left" width="258" height="186">
The potential deformation of the shaft section with length dz will be:
Since during torsion τ = (MK / IP) r, then
Reducing by IP, we obtain an expression for the potential energy of deformation during torsion
4.6 . Torsion of rods of non-circular cross-section
https://pandia.ru/text/78/579/images/image018_20.gif" align="left" width="324" height="237 src="> When torsion of rods (shafts) not round and not annular cross sections, the assumptions accepted for torsion of round and annular shafts are not met: flat cross sections of the rod do not remain flat during torsion, but deplane (curve); straight radii drawn in flat sections are bent; the distance between the sections changes (Fig. 4If a rod of constant cross-section along its entire length is not pinched anywhere and the twisting moments are located at its ends, then all sections deplane equally, and normal stresses do not arise. However, with sufficient accuracy for practical purposes, it can be used for non-round rods. formulas derived for a round rod, replacing both https://pandia.ru/text/78/579/images/image021_17.gif" width="23" height="27 src=">- moment of inertia during torsion, and - moment of resistance during torsion.
https://pandia.ru/text/78/579/images/image024_18.gif" width="90" height="49">, ,
For a rectangular cross section (Fig. 4.12)
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Here and - depend on the relationship.
Coefficients. | The ratio of the larger side of the section to the smaller. |
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Differential" href="/text/category/differentcial/" rel="bookmark">differential equation, the same as the problem of the equilibrium of a thin film stretched over a contour of the same outline as the contour of the cross section of the rod and loaded with uniformly distributed pressure. Analogue voltage is the angle made by the tangent to the film surface with the contour plane, and the analogue of the torque is the volume enclosed between the contour plane and the film surface. Figure 4.13a shows the behavior of the film under pressure; Fig. 4.13b shows the qualitative distribution. stresses during torsion of a rod of complex profile. Using a special device and calibrated film, quantitative results can be obtained. To do this, in order to take into account the rigidity of the film, the same experiment is carried out with a round hole, from where the required film rigidity is obtained, since the solution in this case is possible. get exactly.
4.7. Free torsion of thin-walled rods
Thin-walled rods are those that have one cross-sectional dimension - profile thickness , and less than another - cross-sectional contour length s. Rods come in open (Fig. 4.14) and closed (Fig. 4.15) profiles. Let's use the membrane analogy. The nature of the behavior of the film and, accordingly, the tangential stresses in thin-walled rods of open and closed profiles is fundamentally different (Fig. 4.16 and Fig. 4. If the rod of an open profile is straightened into a long rectangle, then the shape of the film will not change.
Then for a rectangular section at , we have: ,..gif" width="22" height="25"> rectangles, then
..gif" width="42" height="26"> 
.
IN TORSION (Task No. 11)
The task
A steel shaft of circular cross-section consists of three sections with different polar moments of inertia (Fig. 3.6, A). The ends of the shaft are rigidly secured against rotation relative to the longitudinal axis of the shaft. Loads are specified: pairs of forces and , acting in the plane of the cross-section of the shaft; the relationship between the polar moments of inertia of the shaft sections and ; lengths of sections , , .
Required:
1) build a diagram of torques;
2) select the dimensions of the cross sections based on the strength conditions;
3) construct a diagram of the twist angles.
Solution
Due to the presence of two rigid support fastenings, under the influence of load, reactive pairs and arise in each of them. Having created the equilibrium condition for the shaft
We are convinced that the written equation cannot be solved uniquely, since it contains two unknown quantities: and . The remaining equilibrium equations for a given load are carried out identically. Consequently, the problem is once statically indeterminate.
To reveal static indetermination, we create a condition for the compatibility of deformations. Due to the rigidity of the supporting fastenings, the end sections of the shaft do not rotate. This is equivalent to the fact that the total angle of rotation of the shaft in the area A–B equal to zero: , or 
.
The last equation is the condition for compatibility of deformations. To connect it with the equilibrium equation, we write down the physical equations relating torques and angles of twist (3.3) (Hooke’s law for torsion) for each section of the rod:
, 
, 
.
Substituting the physical relations into the condition of compatibility of deformations, we find the reactive moment , and then from the equilibrium equation we determine . The torque diagram is shown in Fig. 3.6, b.
To solve the problem of selecting a section, we write down formulas for determining the maximum tangential stresses (3.5) on each section of the shaft:
; 
; 
.
The coefficients and , representing the ratio of the polar moments of resistance of the sections of the second and third sections of the shaft to the polar moment of resistance of the section of the first section, will be determined through the known parameters and .
The polar moment of inertia can be written in two ways:
; 
,
where , are the radii of the first and second sections of the rod. From here we express the radius through:
Then the polar moment of resistance of the second section
,
that is . Likewise.
Now we can compare the maximum tangential stresses in individual sections and write down the strength condition (3.13) for the largest of them. From this condition we find the required polar moment of resistance, and then, using formula (3.8), the radii of the shaft in each section.
; ; .
To construct a diagram of twist angles, we calculate the twist angles at each section of the rod using formula (3.3). The ordinates of the diagram are obtained by sequentially summing the results for individual sections, starting from one of the ends of the shaft. The correctness of the solution is checked by the equality of the twist angle to zero at the other end of the shaft. The diagram of the twist angles is shown in Fig. 3.6, V.
For a structure with a rigid rod, the rational equilibrium equation, which includes one unknown force, is the equation where A- a hinge around which a rigid rod rotates.
As the name implies, this method is applicable to structures whose rods are made of plastic material.
Obviously, the relationship between the deformations of the rods will be the same as in the first part of the problem, therefore the equation for the compatibility of deformations in the third part of the problem can be written using the previously obtained equation, replacing it with .
When solving this problem, correspondence students perform only calculations based on the plastic limit state. The remaining students solve problem No. 6 in accordance with the teacher’s requirement. Point 2, marked *, is optional and is carried out at the request of the student.
Modern building design standards provide for a more complex approach (introduction of separate safety factors for loads, material properties, operating conditions of the structure). The student will become familiar with this when studying courses on metal, reinforced concrete and other structures.
Let us consider the calculation of a statically indeterminate system for torsion using a specific example.
Example. Based on strength calculations, determine the permissible value of the twisting moment for a shaft rigidly clamped at both ends and loaded, as shown in Fig. 10.8, a.
Rice. 10.8. Scheme of a statically indeterminate shaft
Reactive moments arise in the seals m A And m B(Fig. 10.8, a). Let's create an equilibrium equation relative to the longitudinal axis of the shaft:
Thus, the problem is once statically indeterminate - one static equation and two unknown reactive moments. To compile the displacement equation, we will discard the right seal, replacing its action on the shaft with an unknown reactive torque. The statically determinate system obtained in this way (Fig. 10.8, b) is equivalent to the given one, and, therefore, the angle of rotation of the section B equal to zero:
Applying the principle of independence of the action of forces, we write the displacement equation for the section B as
With only moment acting m 1 section rotation angle IN equal to the angle of twist of the section AC, i.e.
Similarly, under the action of only moment m 2:
With only moment acting m B= X we have:
To simplify the calculations, we express
Substituting the values of rotation angles into the deformation compatibility equation and taking into account the last equality, we obtain:
Substituting the value X into the equilibrium equation, we find:
After revealing static indetermination, diagrams of torques, shear stresses and twist angles are constructed in the usual way. This diagram is presented in Fig. 10.8, c, d, e.
Sections of the area are dangerous BE. It should be noted that the sections in which the greatest torques occur (section AC), in this case less dangerous.
Let us write down the strength condition:
Calculation of cylindrical helical springs
With a small pitch of turns
In many mechanisms and machines, for example, in the springs of carriages and cars, helical springs are used. These springs are wound from wire with a round cross-section made from special grades of steel. When designing such springs, it is necessary to be able to calculate the highest stresses (to check strength) and determine the deformation of the spring (its elongation or deflection). The exact calculation of helical springs is quite complex, since the spring material experiences simultaneously torsion, shear, bending and tension.
A spring with a small pitch of turns is a spring in which the angle between the plane perpendicular to the axis of the spring and the plane of the turn does not exceed 14º.
Let the coil spring be stretched (or compressed) by the forces F, has an average diameter D= 2R and is made of wire with a diameter d(Fig. 10.9, a). To determine the internal forces in a spring, we use the section method.
Rice. 10.9. Diagram of a coil spring with a small pitch of turns
The upper part of the spring (Fig. 10.9, b) will be in equilibrium under the action of an external force F and internal forces in the drawn section of the rod, which can be represented by the sum of the force F and torque M cr.
Assuming that the angle of inclination of the coil is , we can neglect the remaining force factors (longitudinal force and bending moment). The spring material experiences shear and torsion.
The force causes tangential stresses in the cross section, which are determined by the formula
We assume that tangential stresses are distributed evenly over the section (Fig. 10.9, d).
The maximum shear stresses from torsion are:
The distribution of shear stresses due to torsion is shown in Fig. 10.9, c.
The dangerous point on the section contour is the point A, in which the directions of the tangential stresses coincide. Thus, the maximum shear stresses are:
Since in practice, action is often neglected.
Strength condition for springs of small curvature (approximate calculation):
Strength condition for springs of small curvature:
From the formulas for determining it follows that an increase in the diameter of the spring D reduces its strength, and increasing the diameter of the wire d– increases.
When determining the spring deformation, we will take into account only the torsion of the coils.
The change in the length of a spring along its axis under the influence of an external load is called spring draft λ
Where n– number of turns;
G– shear modulus.
Dependence of draft λ from axial load F called characteristics of the spring. Conventional springs have a linear characteristic.
An effort F, at which the movement λ equal to one (1 m), called spring stiffness with, which is determined by the formula
stiffness dimension kN/m.
So, increasing the number of turns n and spring diameter D reduces the spring stiffness, and increasing the wire diameter d increases spring stiffness.
Calculation example
Task 1. For a given circuit (Fig. 10.10, a) determine the value m and build a diagram of torques.
Solution.
1. Using the section method, we determine the value of the torque on each section of the shaft.
Rice. 10.10. Shaft diagram for plotting
torque
1st section. Let us consider the equilibrium of the left cut-off part of the shaft. Let's create an equilibrium equation:
In the 1st section it has a negative value, since from the side of the external normal to the cut off part it rotates clockwise.
2nd section:
3rd section:
on the other side:
2. The diagram taking into account the signs is plotted in Fig. 10.10, b.
Task 2. Shaft of round cross-section with diameter d= 40 mm twists moments m 1 = 0.6 kNm, m 2 = 1.2 kNm and m 3 = 0.8 kNm (Fig. 10.11, a). Check the strength of the shaft and determine the absolute angle of twist of the end section if [ τ ] = 80 MPa, G= = 8×10 4 MPa.
Rice. 10.11. Diagram of a round cross-section shaft
Solution.
1. Using the method of sections, we construct a diagram of torques (Fig. 10.11, b).
2. Let us determine the geometric characteristics of the cross section of the shaft:
3. Check the strength of the shaft:
4. Determine the absolute angle of twist of the end section:
Task 3. Construct a diagram of twist angles for a stepped steel shaft loaded as shown in Fig. 10.12, a. G= = 8×10 4 MPa.
Solution.
1. Let us determine the geometric characteristics of the sections of each shaft stage:
Rice. 10.12. Stepped shaft diagram
2. Let's build a diagram of torques (Fig. 10.12, b).
3. Let us determine the angles of twist in the characteristic sections of the shaft using the formula
Shown in Fig. 10.12, at.
Task 4. Determine the inner and outer diameters of a hollow steel shaft transmitting power N= 100 kW and rotating at angular velocity ω = 80 rad/s if [ τ ] = 60 MPa; α = d/D = 0.6, [ θ ] = 45×10 –4 rad/m, G= 8×10 4 MPa.
Solution.
1. Determine the torque on the shaft:
2. From the strength condition, we determine the moment of resistance of the shaft section:
3. Determine the outer diameter of the shaft from the rigidity condition:
4. We accept D= 80 mm, d= 48 mm.
Task 5. Shaft with a diameter of 90 mm transmits power N= 80 kW. Determine the maximum number of shaft revolutions if [ τ ] = 60 MPa.
Solution.
1. Determine the moment of resistance of the cross section of the shaft:
2. From the strength condition, we determine the torque on the shaft:
3. Determine the maximum number of shaft revolutions:
Task 6. How much power can a shaft rotating with angular power transmit? ω
= 20 rad/s, diameter d= 100 mm at permissible stress [ τ
] = 60 MPa and permissible angle of twist
[θ
] = 45 × 10 –4 rad/m. Shear modulus G= 8 × 10 4 MPa.
Solution.
1. Let us determine the geometric characteristics of the cross section of the shaft:
2. The power transmitted by the shaft is determined by the formula
3. From the strength condition we have:
From the rigidity condition:
We accept
Task 7. d= 6 mm, has n= 10 turns. Spring diameter D= 66 mm. The spring is stretched by axial forces F= 300 N. Check the spring strength if [ τ
] = 240 MPa. Determine the elongation and stiffness of the spring and the stored potential energy.
G= 8 × 10 4 MPa.
Solution.
1. Let's determine the correction factor:
2. Let us determine the tangential stresses arising in the cross sections of the spring rod:
3. Let us determine the elongation of the spring under the action of an external load:
4. Determine the spring stiffness:
5. Let's determine the potential energy of deformation:
Task 8. A steel coil spring is compressed by axial forces F(Fig. 10.13). The spring is wound from wire with a diameter d= 5 mm, in increments t= 10 mm and has a diameter D= 30 mm. Determine the value of the forces F, at which its maximum draft will be reached. G= 8 × 10 4 MPa.
Rice. 10.13. Diagram of a spring compressed by axial forces
Solution.
1. Let us write down the condition for the stiffness of the spring during compression:
2. Determine the limiting value of the forces F:
10.7. Problems to solve independently
Task 9. For a given shaft loading scheme (Fig. 10.14), construct a torque diagram.
Rice. 10.14. Shaft diagram for plotting torque diagrams
Problem 10. For a given stepped shaft (Fig. 10.15), construct a torque diagram.
Rice. 10.15. Scheme of a stepped shaft for plotting
torque
Problem 11. To a constant-section shaft with a diameter d= 50 mm (Fig. 10.16) moments applied m 1 = 1 kNm, m 2 = 0.2 kNm, m 3 = 0.4 kNm and m 4 = 0.4 kNm. Check the strength of the shaft if [ τ ] = 60 MPa. Determine the total angle of twist if G= 8 × 10 4 MPa.
Rice. 10.16. Diagram of a shaft of constant cross-section
Problem 12. Coil spring made of steel wire with diameter d= 5 mm, stretched by forces F= 400 N. Spring diameter D= 30 mm. Check the spring strength if [ τ ] = 500 MPa. Determine the number of turns of the spring at which it lengthens by 40 mm. G= 8 × 10 4 MPa.
Problem 13. Determine the required wire diameter of a helical coil spring for axial load F= 1.2 kN, if D/d= 6 and [ τ ] = 500 MPa.
Problem 14. For a given loading scheme (Fig. 10.17), determine the wire diameters for both springs, if D 1 /d 1 = D 2 /d 2 = 8 and [ τ ] = = 600 MPa.
Rice. 10.17. Diagram of a beam suspended on two springs
10.8. Control questions
1. What type of loading is called torsion?
2. What is called a shaft? What is torque?
3. What deformations occur during torsion?
4. What internal force factors arise during torsion?
5. Derive a formula for determining stresses in the cross section of a twisted round shaft.
6. Derive formulas for determining the relative and total angles of twist of a round shaft.
7. What is a torque diagram and how is it constructed?
8. How is shear stress distributed during torsion? What is the stress at the center of the circular cross section?
9. Write a formula to calculate the stress on the shaft surface during torsion. How will the voltage change if the diameter of the shaft doubles?
10. What is the calculation of torsional strength?
11. What is the calculation for torsional rigidity?
12. Why, with the same strength and rigidity, is a shaft with an annular cross-section lighter than a shaft with a solid circular cross-section?
13. How to calculate the potential strain energy accumulated by a shaft during torsion?
14. Calculation of statically indeterminate shafts.
15. Calculation of cylindrical springs with a small pitch of turns. What is draft and spring stiffness, how are they determined?
Complex resistance