S7: Equacions d'equilibri i lleis d'esforços

Strength of Materials and Structures S7. Isostatic structures: definitions & analysis Lecturers: Pavel B. Ryzhakov, Manuel A. Caicedo Academic year 2020 - 2021 Escola d’Enginyeria de Telecomunicaci´ o i Aeroespacial de Castelldefels Universitat Polit` ecnica de Catalunya This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. 0 Content S7.1: Internal forces in mid-plane structures S7.2: Equilibrium equations in straight elements S7.3: Isostatic and hyperstatic structures S7.4: Internal forces diagrams Exercises 7 Theoretical questions 7 1 S7.1: Internal forces in mid-plane structures S7.1: Internal forces in mid-plane structures Mid-plane structure (estructura de plano medio): Plane structure (straight or curved) in which the straight section is symmetric to a plane containing its longitudinal axis (directriz), being subjected to a system of loads contained in the same plane. If the plane XY is considered as the mid-plane. The following hypotheses can be considered: •R= (Rx,Ry,0) = (N,T,0). Both, shear and axial forces (esfuerzo cortante and esfuerzo axil/axial), will be different from zero. Then, forces acting on zdirection are zero due to symmetry. •M= (0,0,Mz) contains only the bending moment in the zdirection. The components Mtand Myare null due to symmetry. 2 S7.1: Internal forces in mid-plane structures The sign convention is the following: •Axial force N(esfuerzo axil/axial): positive in traction, (positive xon the front side) •Shear force T(esfuerzo cortante): The one that makes the slice rotate counter-clockwise (positive yon the front side) •Bending moment M(momento flector): Positive if it stretches the lower fibre (counter-clockwise moment on the front side of the slice) 3 S7.2: Equilibrium equations in straight elements S7.2: Equilibrium equations in straight elements We consider the equilibrium of a differential slice in a mid-plane straight element subjected to distributed loads (load/ unit length) with components px,py. For equilibrium, we must have (neglecting 2nd order terms): •Balance of forces with respect to x axis: −N+pxdx+(N+dN)=0→px=dN dx •Balance of forces with respect to y axis: −T+pydx+(T+dT)=0→py=dT dx •Balance of moments regarding Gon the front side: −M+Tdx+(M+dM)−py dx2 2= 0 →T=− dM dx These equations define the differential relation between loads p(x) and internal forces (esfuerzos)N(x), T(x) and M(x). In simple cases, we can obtain the internal forces by integrating these equations. 4 S7.3: Isostatic and hyperstatic structures S7.3: Isostatic and hyperstatic structures The forces (loads and reactions) that act on a structure must be in static balance: X i F= 0,X i MO i= 0,for Oarbitrary In the case of loaded structures on the mid-plane, it implies: X i (Fx)i= 0; X i (Fy)i= 0; X i (Mz)O i= 0 These conditions are necessary for ensuring equilibrium but not sufficient. However, for one type of structures they are also sufficient. Isostatic/Hyperstatic Structures To solve a structure means to compute the value of the internal forces (esfuerzos) at its every cross-section. When this can be done by using the available static equations, the structure is called isostatic. Otherwise, it is called hyperstatic. 5 Exercises 7 In order to obtain the equations for the evolution of the internal forces N(x), T(x) and M(x), the equilibrium of external and internal forces acting on the structure, at each segment, must be stated. Segment AB (0 <x<2) On this segment (AB) only acts, as external force, the vertical reaction in A,VA. Therefore, the evolution equations for axial, shear force, and bending moment, as function of x, starting from A, are: N(x)=0,T+ 10 = 0 −→ T=−10 kN M−10 ·x= 0 −→ M(x) = 10 ·x−→ x= 0 −→ M= 0 x= 2 −→ M= 20 kN m 10 Exercises 7 Segment BD (2 <x<6) We proceed similar to the previous segment, but taking into account the augmented system of external forces and reactions acting on the new segment: N(x)=0 T+ 10 −10 = 0 −→ T= 0kN M−10 ·x+ 10(x−2) = 0 −→ M(x) = 20 kN m 11 Exercises 7 Segment DE (6 <x<8) N(x)=0 T+ 10 −10 −10 = 0 −→ T=−10 kN M−10 ·x+ 10(x−2) + 10(x−6) = 0 −→ M(x) = −10 ·x+ 80 x= 6 −→ M= 20 kN m x= 8 −→ M= 0 12 Theoretical questions 7 Theoretical questions 7 Let us consider a beam of length Lsubjected to a bending moment M(x) = x(L−x) (see figure below). We consider a slice of this beam contained in the interval x∈[L/3−δ/2,L/3 + δ/2]. What will be the value of the shear force? 13