P10.Calculate the maximum shear stress due to moment alone.a.226.92 MPa49.50 MPab.221.46 MPad.261.95 MPa11.Determine the maximum shear stress.a.226.92 MPaс.49.50 MPa2.Find the...


STEEL DESIGN<br>BOLTED CONNECTIONS<br>6.<br>Determine the value of P based on net area using a strength<br>reduction coefficient of 85%.<br>Situation 1:<br>For the figure shown, use P = 100 kN, L = 300mm,<br>Si = 100 mm and S2 = 180 mm. Assume the<br>diameter of a bolt is 25 mm.<br>a.<br>b.<br>140.14 kN<br>304.00 kN<br>157.42 kN<br>137.79 kN<br>C.<br>d.<br>Find the value of P based on block shear in gusset plate<br>along the weld.<br>7.<br>157.42 kN<br>137.79 kN<br>a.<br>140.14 kN<br>С.<br>b.<br>304.00 kN<br>d.<br>Compute the value of P based on shear in the weld if D =<br>6mm.<br>8.<br>140.14 kN<br>304.00 kN<br>157.42 kN<br>137.79 kN<br>a.<br>С.<br>S,<br>b.<br>d.<br>4<br>S,<br>The plate shown is connected at the top and at<br>the bottom using 8 mm fillet welds. Use Li = 200<br>mm, L2 = 250 mm, L3 = 300 mm and P = 140 kN.<br>2.<br>Situation 4:<br>S,<br>3.<br>P<br>S,<br>L2<br>L3<br>S2<br>1.<br>Determine the maximum shear stress.<br>a.<br>118.24 MPa<br>С.<br>98.82 MPa<br>33.95 MPa<br>d.<br>92.80 MPa<br>WELDED CONNECTIONS<br>An angle 150 x 100 x 10 mm is connected to the<br>gusset plate as shown. The size of the weld is 8<br>mm. Use P = 400 kN, c =35 mm and E70 electrodes<br>Situation 2:<br>for the welds.<br>9.<br>Solve the shear stress due to force alone.<br>a.<br>226.92 MPa<br>с.<br>49.50 MPa<br>b.<br>221.46 MPa<br>d.<br>261.95 MPa<br>>P<br>10.<br>Calculate the maximum shear stress due to moment alone.<br>a.<br>226.92 MPa<br>49.50 MPa<br>b.<br>221.46 MPa<br>d.<br>261.95 MPa<br>11.<br>Determine the maximum shear stress.<br>a.<br>226.92 MPa<br>с.<br>49.50 MPa<br>2.<br>Find the allowable shear stress in the welds.<br>b.<br>221.46 MPa<br>d.<br>261.95 MPa<br>483.0 MPa<br>C.<br>144.9 MPa<br>b.<br>120.0 MPa<br>d.<br>241.5 MPa<br>3.<br>Compute the total length of weld required.<br>338.07 mm<br>589.34 mm<br>a.<br>С.<br>488.07 mm<br>b.<br>d.<br>292.84 mm<br>4.<br>Solve the length of weld required at the top of the angle.<br>113.88 mm<br>a.<br>38.88 mm<br>C.<br>b.<br>299.19 mm<br>d.<br>374.19 mm<br>Situation 3:<br>A76x76 x 6 mm angular section shown is welded<br>to an 8 mm thick gusset plate. Steel is A36 with Fy<br>= 248 MPa and Fu = 400 MPa. The electrode used<br>is E60 with Fu = 414 MPa. The length Li is 65 mm, L2<br>is 125 mm, and the cross sectional area of the<br>angle is 926 mm2.<br>Gusset<br>Plate<br>L2<br>Allowable stresses:<br>Tension on gross area, Ft = 0.60 Fy<br>Tension on net area, Ft = 0.50 Fu<br>Shear on net and effective area, Fv = 0.30 Fu<br>Shear on welds, Fv = 0.30 Fu<br>5.<br>Calculate the value of P based on gross area.<br>a.<br>140.14 kN<br>с.<br>157.42 kN<br>b.<br>304.00 kN<br>d.<br>137.79 kN<br>

Extracted text: STEEL DESIGN BOLTED CONNECTIONS 6. Determine the value of P based on net area using a strength reduction coefficient of 85%. Situation 1: For the figure shown, use P = 100 kN, L = 300mm, Si = 100 mm and S2 = 180 mm. Assume the diameter of a bolt is 25 mm. a. b. 140.14 kN 304.00 kN 157.42 kN 137.79 kN C. d. Find the value of P based on block shear in gusset plate along the weld. 7. 157.42 kN 137.79 kN a. 140.14 kN С. b. 304.00 kN d. Compute the value of P based on shear in the weld if D = 6mm. 8. 140.14 kN 304.00 kN 157.42 kN 137.79 kN a. С. S, b. d. 4 S, The plate shown is connected at the top and at the bottom using 8 mm fillet welds. Use Li = 200 mm, L2 = 250 mm, L3 = 300 mm and P = 140 kN. 2. Situation 4: S, 3. P S, L2 L3 S2 1. Determine the maximum shear stress. a. 118.24 MPa С. 98.82 MPa 33.95 MPa d. 92.80 MPa WELDED CONNECTIONS An angle 150 x 100 x 10 mm is connected to the gusset plate as shown. The size of the weld is 8 mm. Use P = 400 kN, c =35 mm and E70 electrodes Situation 2: for the welds. 9. Solve the shear stress due to force alone. a. 226.92 MPa с. 49.50 MPa b. 221.46 MPa d. 261.95 MPa >P 10. Calculate the maximum shear stress due to moment alone. a. 226.92 MPa 49.50 MPa b. 221.46 MPa d. 261.95 MPa 11. Determine the maximum shear stress. a. 226.92 MPa с. 49.50 MPa 2. Find the allowable shear stress in the welds. b. 221.46 MPa d. 261.95 MPa 483.0 MPa C. 144.9 MPa b. 120.0 MPa d. 241.5 MPa 3. Compute the total length of weld required. 338.07 mm 589.34 mm a. С. 488.07 mm b. d. 292.84 mm 4. Solve the length of weld required at the top of the angle. 113.88 mm a. 38.88 mm C. b. 299.19 mm d. 374.19 mm Situation 3: A76x76 x 6 mm angular section shown is welded to an 8 mm thick gusset plate. Steel is A36 with Fy = 248 MPa and Fu = 400 MPa. The electrode used is E60 with Fu = 414 MPa. The length Li is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 926 mm2. Gusset Plate L2 Allowable stresses: Tension on gross area, Ft = 0.60 Fy Tension on net area, Ft = 0.50 Fu Shear on net and effective area, Fv = 0.30 Fu Shear on welds, Fv = 0.30 Fu 5. Calculate the value of P based on gross area. a. 140.14 kN с. 157.42 kN b. 304.00 kN d. 137.79 kN
Jun 10, 2022
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