Title Weekly Exam 9 (Steel1 & 2, Hydraulics 1).pdf ✅

F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER 2 ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila WEEKLY EXAM 9 Dream Big, Pray Big, Believe Big. NSTRUCTION: Select the correct answer for each of the following questions. Mark only one answer for each item by shading the box corresponding to the letter of your choice. STEEL DESIGN A bolted lap joint is shown below. The bolts are 20 mm in diameter in 23 mm holes. The plates are 12 mm thick. X1 = 33 mm, X2 = 76 mm, X3 = 50 mm. Yield strength of plate Fy = 248 MPa. Ultimate tensile strength of plate, Fu = 400 MPa. 1. Find the allowable load P based on gross area yielding A. 430 kN B. 521 kN C. 450 kN D. 489 kN 2. Find the design load P on the net area rupture A. 658 kN B. 574 kN C. 439 kN D. 498 kN 3. Find the design load P based on block shear A. 795 kN B. 644 kN C. 530 kN D. 768 kN The line of action of the force P coincides with the axis of the angle which is at a distance a = 29 mm from the back of the connected angle. Properties of one angle 100 mm x 100 mm x 8 mm: A = 1770 mm2 Fy = 248 MPa Allowable weld shear stress, Fvw = 93 MPa 4. Determine the maximum tensile capacity of the angle A. 236 kN B. 362 kN C. 263 kN D. 326 kN 5. Find the value of L2 using 6 mm fillet weld A. 474 mm B. 193 mm C. 447 mm D. 139 mm 6. Determine the value of L2 when a weld size of 5 mm is added at the end of the angle A. 295 mm B. 432 mm C. 423 mm D. 295 mm Two channels are joined together by eight 25 mm diameter bolts in a single shear as shown in the figure. The bolts spacing are x = 50 mm and y = 60 mm. The steel has Fy = 250 MPa and Fu = 400 MPa Channel properties: A = 3644 mm2 Web thickness = 12 mm Bolt hole diameter = 27 mm Nominal shear strength of bolt when threads are not excluded from the shear plane = 0.45 Fu 7. Find the design load P based on bolt shear A. 353 kN B. 350 kN C. 530 kN D. 503 kN 8. Find the design load P based on bearing on plate A. 1802 kN B. 1265 kN C. 1152 kN D. 1611 kN 9. Find the allowable load P based on block shear A. 658 kN B. 734 kN C. 652 kN D. 749 kN A W 10 x 45 is used as a beam with two lines of 19 mm diameter bolts in each flange using A-36 steel. The diameter of a hole is 3 mm greater than the bolt diameter. There are at least three bolts in each line and the bolts are not staggered with respect to each other. Use LRFD method. Properties of W 10 x 45: A = 8581 mm2 tw = 8.89 mm d = 256.54 mm Fy = 248 MPa bf = 203.71 mm Fu = 400 MPa tf = 15.75 mm 10. Find the effective net area that can be assumed to resist tension of the section through the holes. U = 0.90 A. 7195 mm2 B. 6475.5 mm2 C. 9571 mm2 D. 7465.5 mm2 11. Find the design strength due to fracture in the net section where bolt or rivet holes are present A. 1942.65 kN B. 1694.52 kN C. 1426.59 kN D. 1529.46 kN
F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER 2 ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila WEEKLY EXAM 9 Dream Big, Pray Big, Believe Big. 12. Find the design strength due to yielding in the gross section which is intended to prevent excessive elongation of the member. A. 1591.82 kN B. 1812.59 kN C. 1915.28 kN D. 1298.51 kN The beam connection shown in the figure is composed of two plates (both sides) bolted to web of the beams. In this problem, e = 200 mm, x = 75 mm, y = 65 mm. 13. Determine the vertical component of the force on the most critical bolt due to direct shear A. 40 kN B. 25 kN C. 20 kN D. 30 kN 14. Find the horizontal component of force on the most critical bolt due to torsional shear A. 56.32 kN B. 48.75 kN C. 69.55 kN D. 20.09 kN 15. Find the total shear stress on the most critical bolt if the diameter of the bolt is 20 mm A. 148 MPa B. 156 MPa C. 165 MPa D. 132 MPa Two channels are welded at the tip of their flanges form a built-up column as shown. To strengthen the column, cover plates are added at the top and at the bottom flanges. Use Fy = 248 MPa Given: Unsupported column height: 9.2 m Column is braced against sidesway in both directions Column ends are fixed, k = 0.70 Section Properties of the channel: Area, A = 4529 mm2 Depth, d = 229 mm Flange width, bf = 87 mm Flange thickness, tf = 14 mm Web thickness, tw = 10 mm Ix = 35.4 x 106 mm4 Iy = 3.0 x 106 mm4 rx = 88.4 mm ry = 25.6 mm Distance from the back of the web to the y-axis, x = 24.9 mm Modulus of elasticity, Ey = 200 GPa 16. Without cover plates, compute the allowable axial load in the column. A. 874 kN B. 1035 kN C. 1236 kN D. 1052 kN 17. Calculate the critical bucking load if there are no cover plates. A. 1594 kN B. 1580 kN C. 4860 kN D. 3818 kN 18. If cover plates 150 mm wide and 12 mm thick are added, which of the following gives the ultimate axial load? A. 1563 kN B. 2253 kN C. 2103 kN D. 2336 kN 19. If cover plates 150 mm wide and 12 mm thick are added, which of the following gives the critical buckling load? A. 3334 kN B. 4448 kN C. 3844 kN D. 4384 kN Given: Dimension: a = 0.9 m c = 1.8 m b = 0.9 m h = 1.8 m Section properties: A = 640 mm2 r = 20 mm Yield strength of steel, Fy = 248 MPa Modulus of elasticity, E = 200 GPa Effective length factor, k = 1.0
F.A.L. CONDUCIVE ENGINEERING REVIEW CENTER 2 ND Floor, Cartimar Building, C.M. Recto Avenue, Quiapo, Manila WEEKLY EXAM 9 Dream Big, Pray Big, Believe Big. 20. Find the allowable compressive force of member AB A. 56 kN B. 66 kN C. 46 kN D. 36 kN 21. Find the allowable compressive force of member AC A. 36 kN B. 56 kN C. 46 kN D. 66 kN 22. Find the allowable compressive force of member AD A. 63.6 kN B. 52.4 kN C. 40.6 kN D. 35.4 kN 23. Find the safe value of W. A. 86.1 kN B. 72.0 kN C. 36.0 kN D. 81.2 kN A steel frame is shown. The frame is braced against joint translation (sidesway). The girders are W 21 x 402 with I = 0.0051 m4 . The columns are W 21 x 93 with I = 0.0086 m4 . It is required to determine the effective length of column KL using the alignment chart given. The subscripts A and B in the chart refer to the joints at the two ends of the column section. The factor G is defined as G = ∑ ( Ic Lc ) ∑ ( Ig Lg ) in which Σ indicates a summation of all members rigidly connected to that joint. Ic is the moment of inertia and Lc the unsupported length of a column section. Ig is the moment of inertia and Lg the unsupported length of a girder or other restraining member. In calculating G, only rigidly attached girders or columns are included. For column ends supported by but not rigidly connected to a footing or foundation, G is theoretically infinity, but, unless designed as a true fiction free pin, may be taken as “10” for practical designs. If the column end is rigidly attached to a properly designed footing, G may be taken as 1.0 24. Which of the following most nearly gives the factor G for the top end of member KL? A. 2.87 B. 2.14 C. 1.77 D. 1.54 25. Which of the following most nearly gives the factor G for the bottom end of member KL? A. 1.54 B. 2.57 C. 3.21 D. 1.73 Alignment chart – sidesway uninhibited HYDRAULICS 1 A thin plate moves between two parallel horizontal, stationary flat surfaces at a constant velocity of 5 m/s. The two stationary surfaces are spaced 4 cm apart, and the medium between them is filled with oil whose viscosity is 0.9 N-s/m2. The part of the plate immersed in oil at any given time is 2 m long and 0.5 m wide. The plate moves through the mid-plane between the surfaces. 26. Determine the force required to maintain this motion A. 113 N B. 225 N C. 540 N D. 450 N 27. What would the force F be if the plate is 1 cm from the bottom and 3 cm from the surface. A. 600 N B. 450 N C. 300 N D. 900 N