# Chapter 3 THE STATIC ASPECT OF SOLICITATION

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2 The resultant R (Fig. 3.1) is the area of the load diagram, and it is located in the center of gravity (centroid, center of mass) of this diagram. Fig on bars: p [force/length] units: [kn/m], [dan/cm].. - on slabs: p [force/area] units: [kn/m 2 ], [dan/cm 2 ].. - in massifs : p [force/volume] units: [kn/m 3 ], [dan/cm 3 ].. c) concentrated (point) moments: M [force length] 3.2 Supports a) The roller (simple support) - It is free to rotate and translate along the surface upon which the roller rests - The translation perpendicular to the surface upon which the roller rests is prevented (Fig. 3.2) Fig. 3.2 b) The pinned support (hinge) Fig The reaction force R V is always a single force that is perpendicular to the surface upon which the roller rests - It will allow the structural member to rotate, but not to translate in any direction (Fig. 3.3) - The reaction forces R V and R H are always two forces acting perpendicular, respectively tangent to the surface upon which the hinge rests.

3 c) The fixed (built-in) support - A fixed support does not permit rotation and translations in any direction - There are 3 reactions: vertical and horizontal reactions R V and R H and a fixed end moment M (Fig. 3.4) Fig. 3.4 All construction elements should be in equilibrium under the external actions and reactions from supports. In plan we may write 3 equations of equilibrium: - Vertical equilibrium: total forces acting upward = total forces acting downward - Horizontal equilibrium: total forces acting to the right = total forces acting to the left - Moment equilibrium: total moments rotating clockwise = total moments rotating anticlockwise Having three available equations of equilibrium, we may solve a system which has up to three unknown reactions. That s why from statically point of view we met: - statically determinate structures (isostatic): Simply supported beam Cantilever Simple supported beam with free ends (overhanging beams) Simply supported frame

4 The number of unknown reactions is exactly equal to the number of available equations of equilibrium, three ( = 0 ; = 0 ; = 0 ) - statically indeterminate structures (hyperstatic): the number of the reactions from supports is greater than the number of the equilibrium equations. - unstable structures: the number of the reactions from supports is smaller than the number of the equations of equilibrium. 3.3 REACTIONS Considering the hypothesis of the small deformations, the equations of equilibrium may be written on the initial undistorted, undeformed member, acting like a rigid body (Fig. 3.5). The 3 equations which will be used are: = 0 ; = 0 ; = 0 Fig. 3.5

5 = 0 = 0: = 0 => = (Forces acting to the right are positive) = 0 = 0 => = (The clockwise moment is arbitrarily taken as positive) = 0 = 0 => = Verification: = 0 = 0: + = 0 (Forces acting downward are positive) 3.4 THE STATIC DEFINITIONS OF INTERNAL STRESSES (INTERNAL CONCENTRATED FORCES AND MOMENTS) We consider a beam in static equilibrium under the action of external forces and reactions. Left part (I) Right part (II) Fig. 3.6

6 We may find the internal stresses, using the method of sections. We cut the bar (Fig. 3.6) with a section n-n perpendicular to the longitudinal bar axis Gx, obtaining 2 separate parts: left part (I) and right part (II). Assuming the left part is removed, to assure the equilibrium of the right part, on the face of this part we have to introduce internal stresses that replace the action of the removed part (left). These internal stresses are reduced in the cross section centroid (of the right part) to: - an axial force N (tangent to the longitudinal bar axis) - a shear force V z (perpendicular to the longitudinal bar axis) - a bending moment If instead the left part I, the right part II is removed, the internal stresses are introduced on the left part in the same manner, but opposite to the first case. The internal stresses were introduced on each face with the positive convention. Let s define these stresses and see their positive signs: Axial force N: is the algebraic sum of all forces acting longitudinal (parallel) to the bar axis, taken from the left or the right side, from the section considered. It is positive when it produces tension in the cross section. Shear force V: is the algebraic sum of all forces acting transverse (perpendicular) to the bar axis, taken from the left or the right side, from the section considered. It is positive when on the left it acts upward and contrary. Bending moment M: is the algebraic sum of the moments of all forces taken from the left or the right side, from the section considered, taken about the centroid of the cross section. It is positive when it produces tension in the bottom fiber of the cross section.

7 3.5. DIAGRAMS OF STRESSES Each cross section is positioned along the bar by the variable x. The internal stresses are also functions of this variable x. The diagrams of stresses are even the graphs of these functions. To draw these diagrams it is enough to write the analytical expression of N(x), V z (x) and M y (x) and then to represent the graphs of these functions. But, this procedure must be applied on different intervals, where external loads or reactions act. Example: A-B: 0 1 N(x)=0 ( ) = = 2 = = B-C: 1 2 N(x)=0 = 1 = 2 = 1 0,5 = C-D: 2 3 D-E: 3 4 N(x)=0 = 1 = 2 = 0,5 + = 2 + 3,4 ( ) = = 12 = 1 + = 4 = 0, = 4 14,6

8 3.6. RELATIONSHIPS BETWEEN LOADS AND STRESSES We consider a differential element dx along the length of a beam (Fig.3.7). The ends of the element are subject to shear forces and moments which are represented with their positive directions. On this differential element a uniformly distributed load p n acts, considered positive when it acts downward. We write the vertical equilibrium: V + p d + V + dv = 0 Fig.3.7 = (3.1) The above relation (3.1) represents the first differential relationship between the shear force and load and it can be enunciated as: the rate of change of shear force, which is the slope of the shear diagram, is minus the intensity of the distributed load acting transverse to x axis. Similarly (Fig.3.8), from horizontal equilibrium: N + p d + N + dn = 0 Fig.3.8 = (3.2) This relation (3.2) represents the second differential relationship between the axial force and load and it can be enunciated as: the rate of change (the first derivate) of axial force, which is the slope of the axial force diagram, is minus the intensity of the distributed load acting along x axis. Writing now the moment equilibrium about the centroid of the differential element (Fig.3.7):

9 M + V 2 + V dx 2 + dv dx 2 M dm = 0 Neglecting the second order term, as dx 0 => = => = (3.3) This final relation (3.3) represents the third differential relationship between the bending moment and shear force and it can be enunciated as: the rate of moment, which is the slope of the moment diagram, is equal to the shear force. These relations (first and last) may be written also: = = (3.4) To understand better these differential relations we may say that if ( ) is a polynomial of degree n, ( ) is a polynomial of degree n+1, respectively ( ) is a polynomial of degree n+2 (Fig.3.9) UNIT STRESSES Fig.3.9 In the previous paragraphs we have seen that to determine the internal forces, the procedure to separate parts of the body, to exteriorize internal forces and to write equilibrium equations, are the main procedure in Mechanics of Materials. The internal forces in the sectioned plane represent the resultant of small contact forces that were released due to sectioning of that element (Fig.3.10).

10 Fig.3.10 If an infinitely small area da (Fig.3.11) is considered on the sectioned surface and the resultant force d is calculated for this area, it can be assumed that: (1) d is applied in the centroid of da (2) if da is very small, d can be considered uniformly distributed on da. The distributed force on da is: Fig.3.11 = (3.5) The new notion is called unit stress. If da=1=> = d, so the unit stress is the interior force uniformly transmitted through the unit surface (area). It is measured in units:,,.

11 The unit stress is a vectorial notion, which can be decomposed in 2 main components (Fig.3.11): - a normal component, called normal unit stress or direct stress - a tangent component, called tangential unit stress or shear stress We observe that = + and both components have a first index which indicate the direction of the axis which is perpendicular to the cross section. The tangential stress can be decomposed (Fig.3.11) into 2 components: parallel to Gz axis and parallel to Gy axis. The orientation of and from figure (Fig.3.11) is the positive one: is positive if it comes from the cross section. 3.8 THE STRENGTH CALCULATION OF STRESSES (THE REDUCTION OF THE UNIT STRESSES IN THE CROSS SECTION) Fig.3.12 The resultant of the unit stress p on the differential area da, p da, must be reduced into two significant points of the cross section (Fig.3.12): - the center of gravity (centroid) G, where the resultant of the normal stress σ x da will be reduced - the shear center C, where the resultant of the tangential stress τ x da will be reduced

12 3.8.1 The reduction of the normal stress σ x Reducing (σ x da ) in G (Fig.3.13), we obtain: - an elementary force: = - an elementary moment: Fig.3.13 = y = ρ cosα z = ρ sinα The elementary moment dm i is decomposed into: = = = = So, the elementary components of the moment dm i, about Gy and Gz axis are: = = Extending all the elementary forces dn on the entire area A, we obtain finally the axial force: = (3.6) Similarly, the elementary moments give the bending moments: = = (3.7) (3.8)

13 !!! The bending moments have the vectors orientated along Gy and Gz axis (perpendicular to the longitudinal bar axis Gx). The positive orientation (Fig.3.13) of the axial force N and the bending moments M y and M z on the cross section is: The reduction of the tangential stress τ x Fig.3.14 Reducing (τ x da) in C (Fig.3.15), through its components (τ xz da) (τ xy da), we obtain: - the elementary forces: = = - the elementary moment: Fig.3.15 = = Extending on the entire area A, we obtain the shear forces: = = (3.9) (3.10) Similarly, the elementary moment give the torsion moment (torque): = [ (y y ) (z z )]da (3.11)!!!The torque M t has the vector orientated along axis (the longitudinal axis of the shear center).

14 The positive orientation (Fig.3.15) of the shear forces V z and V y, and the torsion moment M t on the cross section is: Fig.3.16 The internal stresses written in this way, from strength calculation is called the stress calculation from interior, while the static calculation is called calculation from exterior BARS ACTIONS (SOLICITATIONS) a) Simple actions (when a single stress acts in the cross section) - centric (axial) tension : N>0, (V z = V y =0, M y = M z = M t =0) - centric (axial) compression : N<0, (V z = V y =0, M y = M z = M t =0) - pure shearing : V z 0 or V y 0, (N =0, M y = M z = M t =0) - pure bending : M y 0 or M z 0, (V z = V y =0, N = M t =0) - pure torsion : M t 0, (V z = V y =0, M y = M z = N =0) b) Compound actions (when more stresses act in the cross section) - eccentric tension or compression : N and M y and/or M z - pure skew (oblique) bending : M y and M z - shearing with torsion: V z and/or V y and M t - oblique bending with shearing : M y and M z, V y and V z - eccentric tension or compression with shearing: N and M y and/or M z, V y and V z - eccentric tension or compression with shearing and torsion: N and M y and/or M z, V y and V z and M t

15 3.10 APPLICATIONS FOR DIAGRAMS OF STRESSES Represent the diagrams of stresses for the simply supported beam from Fig.3.17 Fig.3.17 Reactions: (1) = 0: = 0 (2) = 0: 7 3,5 5 = 0 = 49 kn (3) = 0: 7 1,5 + 5 = 0 = 21 kn Verification: = 0: + 7 = 0 OK Stresses: Writing efforts will be traversing bar from left to right. In the sections i where concentrated forces act (horizontal and vertical), the axial force N and the shear force V will be written in sections located infinitely left i and infinitely right i section, with respect to the section i. Bending moment M will be written directly into the section.

16 For a better understanding of writing efforts in a section, in this application efforts will be written in each section only taking the first left and then the right section, obviously obtaining the same results. Axial force N With no forces tangential or parallel to the longitudinal axis of the bar, axial force is zero throughout the bar. Shear force V,, = = 21 (taken from the left),, = 7 = 21 (taken from the right), = 5 = 29 (taken from the left), = + 2 = 29 (taken from the right),, = 5 + = 20 (taken from the left),, = 2 = 20 (taken from the right) Bending moment M = 0 (taken from the left) = 7 3,5 + 5 = 0 (taken from the right) = 5 2,5 + 5 = 20 (taken from the left) = 2 1 = 20 (taken from the right) In the free end V = M = 0. Diagrams (graphs) of stresses: - N diagram is represented with positive values (tensile stress) above the reference line, and negative values (compression) below the reference line - V diagram is represented with positive values above the reference line, and negative values below the reference line - M diagram is always represented on the tension fiber (for convenience, the positive values are below the reference line, and the negative values are above the reference line)

17 Returning to our problem, with the values obtained the diagrams of stresses are represented. It is noted that V is zero in a section (between A and B) whose position is unknown. Given the differential relationship (3.3) that exists between V and M, in the section where V is zero, M will have an extreme value. To calculate the extreme value, we must first determine where is this section of V=0. We denote the distance from the support A to that section with x, and we write the condition that in that section V (x) = 0. = = 0 = 2,1 With this known distance, the maximum moment is: = = + = 22,05 From the graphs we see again the connection between the distributed load q, V and M, in that V is a high degree (linear variation) with respect to the uniformly distributed load q (constant), and M is with a high degree (parabolic variation) with respect to V, having a maximum section where V is zero. In the section where a concentrated force acts, V diagram will present a sudden jump, in the amount equal to the force Represent the diagrams of stresses for the simply supported beam from Fig.3.18 Reactions: (1) = 0: + = 0 = 200 kn (2) = 0: = 0 = 160 kn (3) = 0: = 0 = 300 kn Verification: = 0: = 0 OK Axial force N,, = = 250 =,,, = = 200 =,,, = = 0

18 Fig.3.18 Shear force V z,, = 0, = 3 = 120,, = 3 + = 180 = 8 + = 20 =,,, = 0 =, Bending moment M y,, = 0 = 3 1,5 = 180 = = 220 = = 180 = Determine the position of the section in which V = 0:

19 = = 0 = 7,5 The maximum moment is: = = + 3 = Represent the diagrams of stresses for the cantilever from Fig.3.19 Fig.3.19 On cantilever we can write the stresses from the free end to the fixed support (in this application, from the right), such that the calculation of the reactions is not necessary. The axial force N is zero in the entire bar. Shear force V = 0,, = 2 = 120, = 2 = 120,, = 6 = 120 Bending moment M = 0 = 2 1 = 120 = = 120 = = 0

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