Imperial


AASHTO-LRFD Shear Strength for superstructure tutorial’s content is transverse reinforcement requirement, critical section for shear, minimum reinforcement, shear stress on concrete, maximum spacing of transverse reinforcement, nominal shear resistance, concrete and steel components, determination of β and  and lastly net tensile strain. Transverse reinforcement which is the first topic is needed if there is a significant chance of diagonal cracking. Where the reaction force in the direction of applied shear introduces compression into the end regime of a member, critical section is distance dv. Concentrated load acts between support and dv away from internal face of the support. As both cases are possible, the critical section for shear will be taken conservatively at the internal face of the support. AASHTO specifies a minimum amount of transverse reinforcement if transverse reinforcement is needed as specified in Article 5.8.2.4. Minimum amount of transverse reinforcement is specified to prevent growth of diagonal cracking and to increase ductility of a section. If the shear stress at design section is greater than 0.18f’c  and the beam type element is not built integrally with supports, section should be design for shear using strut and tie model. The upper limit of nominal shear resistance is to ensure that the concrete in the web will not crush prior to yielding of transverse reinforcement. β and  are determined using two methods. A direct calculation using the formula given in Article 5.8.3.4.2 .A tabularized values presented in Appendix-5 of AASHTO-LRFD. Flexural tension side of a member is half of the section, which contains flexural tension zone. If net tensile strain calculated as stated previously is negative, it should be taken zero or denominator should be changed by . However, net tensile strain can not be less than -0.40 x 10-3. For sections closer than dv to face of support, value of  calculated at dv away from support can be used.  can be taken greater than the calculated value of it but it can not exceed 6.0×10-3 .

Metric


AASHTO-LRFD Shear Strength for superstructure tutorial’s content is transverse reinforcement requirement, critical section for shear, minimum reinforcement, shear stress on concrete, maximum spacing of transverse reinforcement, nominal shear resistance, concrete and steel components, determination of β and  and lastly net tensile strain. Transverse reinforcement which is the first topic is needed if there is a significant chance of diagonal cracking. Where the reaction force in the direction of applied shear introduces compression into the end regime of a member, critical section is distance dv. Concentrated load acts between support and dv away from internal face of the support. As both cases are possible, the critical section for shear will be taken conservatively at the internal face of the support. AASHTO specifies a minimum amount of transverse reinforcement if transverse reinforcement is needed as specified in Article 5.8.2.4. Minimum amount of transverse reinforcement is specified to prevent growth of diagonal cracking and to increase ductility of a section. If the shear stress at design section is greater than 0.18f’c  and the beam type element is not built integrally with supports, section should be design for shear using strut and tie model. The upper limit of nominal shear resistance is to ensure that the concrete in the web will not crush prior to yielding of transverse reinforcement. β and  are determined using two methods. A direct calculation using the formula given in Article 5.8.3.4.2 .A tabularized values presented in Appendix-5 of AASHTO-LRFD. Flexural tension side of a member is half of the section, which contains flexural tension zone. If net tensile strain calculated as stated previously is negative, it should be taken zero or denominator should be changed by . However, net tensile strain can not be less than -0.40 x 10-3. For sections closer than dv to face of support, value of  calculated at dv away from support can be used.  can be taken greater than the calculated value of it but it can not exceed 6.0×10-3 .