Explanation Given: Number of bolts are 6 . Steel Used is A992 . Dead load is 30 kips . Live load is 70 kips . Calculation: The properties for A992 steel from the AISC steel table are as follows: The nominal shear stress is 54 ksi . The ultimate tensile stress is 65 ksi . Write the expression for cross-sectional area of the bolt. A b = π 4 d 2 ...... (I) Here, cross-sectional area of the bolt is A b and diameter of the bolt is d . Substitute 3 4 inch for d in Equation (I). A b = π 4 ( 3 4 inch ) 2 = 0.4418 inch 2 Write the expression for nominal shear capacity of one bolt. R nv = ( F nv A b ) ...... (II) Here, nominal shear stress of one bolt is R nv and nominal shear stress of the bolt material is F nv . Substitute 0.4418 inch 2 for A b and 54 ksi for F nv in Equation (II). R nv = ( 54 ksi ) ( 0.4418 inch 2 ) = 23.86 kips Write the expression for the bearing strength of an edge bolt R n . R n = ϕ 2.4 d t F u ...... (IV) Here, bearing strength is R n , minimum thickness of the section is t , load reduction factor is ϕ , and the ultimate tensile stress is F u . Substitute 0.605 inch for t , 3 4 inch for d , 0.75 for ϕ , and 65 ksi for F u in Equation (IV). R n = 0.75 × 2.4 × 3 4 inch × 0.605 inch × 65 ksi = 53.1 kips Write the expression for total factored load. P u = 1.2 D + 1.6 L ...... (V) Here, total factored load is P u , live load is L , and dead load is D . Substitute 70 kips for L and 30 kips for D in Equation (V). P u = 1.2 ( 30 kips ) + 1.6 ( 70 kips ) = 148 kips Write the expression for the tensile component of the load. T u = P u × cos 30 ° ...... (VI) Here, tensile force is T u . Substitute 148 kips for P u in Equation (VI). T u = 148 ksi × cos 30 ° = 128.17 kips Write the expression for shear component of the load. V u = P u × sin 30 ° ...... (VII) Here, the shear force is V u . Substitute 148 kips for P u in Equation (VII). V u = 148 ksi × sin 30 ° = 74 kips Write the expression for tensile force per bolt To determine (b) If the connection is adequate using Allowable Stress Design (ASD) method.