PSI - Issue 78

Pier Paolo Rossi et al. / Procedia Structural Integrity 78 (2026) 726–733

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by surfaces S 2 and S 3 . Regarding surface S 4 , the authors recognize that it can be fairly close or far from the re-entrant corner of the nib. Owing to this, it is impossible to determine, once and for all, which rebars are cut off by this failure surface. To this end, the angle of inclination of S 4 is derived from the equation of equilibrium in the transverse direction of a RC element with longitudinal and transverse reinforcement. This element is assumed to be on the inclined part of surface S 4 and obtained by cutting the full-depth section of the member with two planes, which are parallel and orthogonal to the principal compression stress of concrete, and with a plane that is parallel to the longitudinal member axis. The translational equilibrium of this element along the vertical direction states that 2 sy sy ci sin 0     where  sy is the stress of the stirrups,  ci is the principal compression stress of concrete and  is the angle of inclination of the principal compression stress of concrete with respect to the longitudinal axis of the members. The geometric ratio of the transverse reinforcement sy sy   A bs , where b is the width of the beam, s is the spacing of the stirrups and A sy is the projection of the cross-sectional area of the transverse reinforcement in the direction of the shear force. To calculate the value of the angle  the normal stress  ci is first fixed equal to the concrete strength under biaxial stress state and the normal stress  sy is assigned equal to the stress of the transverse reinforcement. Then, the cotangent of the angle  is compared with limit values of the same parameter. The lower and upper limit values of the cotangent of the angle  are calculated supposing that the failure crack angle may be 20–25° lower or higher than the angle of inclination  I of the first crack. In the absence of external axial loads (  I = 45°) and assuming a variation of the angle  equal to 23.2°, the limit values of the cotangent of the angle  are 0.4 and 2.5. Additionally, another upper limit is assumed for the cotangent of the angle  , which is the one corresponding to the inclined part of the failure surface that intersects the hanger (the one closer to the loading plate) at a vertical distance from the bottom of the full-depth section equal to the concrete cover. If the obtained value of cotg  is higher than the upper limit value, the value of cotg  is fixed equal to the higher limit and the stress of concrete is derived. Instead, if cotg  is lower than the lower limit, the solution with the assumed value of the stress in the transverse reinforcement is unacceptable and a constraint is formulated in the optimization problem to force cotg  to be equal or higher than the lower limit. 2.2. Response of concrete and steel rebars With the aim of applying the static theorem of plasticity, the response of concrete and steel rebars is considered to be perfectly plastic and ductile. The uncracked concrete in the compression chord of the beam is assumed to be in a biaxial state of stress and to resist normal stresses combined to shear stresses. The longitudinal and transverse steel bars are assumed to resist uniaxial compressive and tensile stresses. The dowel action of the rebars is neglected. 2.3. Variables of the problem For each of the considered failure surfaces, the variables of the problem are the depth x of the compressed part of the failure surface, the axial stresses of the longitudinal, transverse and diagonal rebars that are cut off by the failure surface, i.e.  sy ,  sz and  si . To limit the number of variables, the stress of the longitudinal rebar contributing to the equilibrium of the part of the beam that is cut off by S 1 is equal to the stress of the same longitudinal rebar in the equilibrium of the part of the beam that is cut off by S 2 , S 3 or S 4 . In addition, the normal stress is equal in all the longitudinal rebars that are at an equal distance from the bottom of the beam and that are characterized by an equal yield strength of steel. In addition, the stress is equal in the hangers that are at an equal distance from that re-entrant corner of the nib and that are characterized by an equal yield strength of steel and equal in all the stirrups with equal spacing and yield strength of steel. 2.4. Equilibrium equations In the general case in which horizontal, vertical and diagonal rebars are present in the dapped-end beam the equations of translational equilibrium in the horizontal and vertical directions (Fig. 2) are

z,i   N N 

cos

0

  N

(1)

dz,i

c