Civil **Text Transcription:**In the connection shown in Figure 5, ASTM A572 Grade 50 steel is used for the tension member (C10×20). A36 steel is used for the gusset plate (t = ½ in.).The holes are for 1-inch bolts. What is the maximum factored block shear strength in LRFD for this connection?**Note:** You must check both, the tension member and the gusset plate, for block shear.---**Diagram Explanation:**- The diagram (Figure 5) illustrates a structural steel connection.- A gusset plate with a thickness of ½ inch is connected to a channel section (C10×20).- It shows the placement of several holes, aligned in two rows, intended for 1-inch bolts.- Key dimensions: - Horizontal distance between the hole centers is 3 inches. - Vertical edge distances from the edge to the center of the first hole are 1½ inches. - Vertical spacing between the rows of holes is 2 inches.

C 10 x 25 ( A = 7.35 in2 , tw = 0.526 in )A572 grade 50 steel (Fy = 50 ksi and Fu = 65 ksi)We know that factored Block shear strength is given by .Rn = .(0.6*fu*Anv+Ubs *fu*Ant) .(0.6*fy*Agv+Ubs*fu*Ant)Agv = gross area subjected to shearAgv = (1.5+3+3)*0.526 * 2 (since 2 shear plane)Agv = 7.89 in2 Anv = Net area subjected to shearAnv = Agv - 2*2.5 *Area of bolt hole (2 shear plane each with 2.5 number of bolt hole)Area of bolt hole = diameter of bolt hole * thickness of bolt holediameter of bolt hole = 17/16 in (Standad bolt hole diameter for 1 inch bolt diameter = 17/16 in from AISC J3.3)thickness of bolt hole = 0.526 inArea of bolt hole = 17/16 * 0.526 = 0.56 in2Anv = 7.89 - 2*2.5*0.56Anv = 5.09 in2 Ant = Net area subjected to tension Ant = (6*0.526 - 1*Area of bolt hole) Ant = (6*0.526 - 1*0.56) Ant = 2.596 in2Ubs =1 for single row of connectionNow substitute the values to get '.Rn'.Rn = 0.75(0.6*65*5.09+1*65*2.596) 0.75(0.6*50*7.89+1*65*2.596).Rn = 275.44 K < 304.08 KMinimum of above two is selected hence .Rn = 275.44 KWe know that factored Block shear strength is given by .Rn = .(0.6*fu*Anv+Ubs *fu*Ant) .(0.6*fy*Agv+Ubs*fu*Ant)Agv = gross area subjected to shearAgv = (3+3+1.5)*0.5 * 2 (since 2 shear plane)Agv = 7.5 in2 Anv = Net area subjected to shearAnv = Agv - 2*2.5 *Area of bolt hole (2 shear plane each with 2.5 number of bolt hole)thickness of bolt hole = 0.5 inArea of bolt hole = 17/16 * 0.5 = 0.53 in2Anv = 7.5 - 2*2.5*0.53Anv = 4.85 in2 Ant = Net area subjected to tension Ant = (6*0.5 - 1*Area of bolt hole) Ant = (6*0.5 - 1*0.53) Ant = 2.47 in2Ubs =1 for single row of connectionNow substitute the values to get '.Rn'.Rn = 0.75(0.6*65*4.85+1*65*2.47) 0.75(0.6*50*7.5+1*65*2.47).Rn = 262.275 K < 289.16 Khence .Rn = 262.275 KMaximum factored Block shear strength of the connection considering tension member and gusset plate is minimum block shear strength given by the twoHence Maximum factored block shear strength is given by gusset plate .Rn = 262.275 K" Maximum factored block shear strength in LRFD for this connection .R,nbsp; .Rn = 262.275 K "