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Republlic of the Philippines Eastern Visayas State University Tacloban City
Set A
College of Engineering Civil Engineering Department MID PRE-BOARD in Structural Design
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In Figure SMAT 1804, assume that a 25-mm-diameter rivet joins the plates that are each 100 mm wide. The allowable stresses are 150 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; a 7.85 mm. b 17.41 mm. c 13.09 mm. d 21.86 mm. and (b) the largest average tensile stress in the plates. a 18 Mpa. b 50 Mpa. c 16.5 Mpa. d 30.06 Mpa. Given the Figure SMAT 1809: Vertical load P = 6000 lb; Crosssectional area of each rod = 0.5 in²; E = 10 × 10⁶ psi; α = 30°; θ = 30°; Required: (a)Deformation ∆ AB = ? a 0.144 in. b -0.1149 in. c -0.0864 in. d -0.1443 in. Deformation ∆ BC = ? a 0.144 in. b -0.0864 in. c -0.0285 in. d 0.2405 in. Horizontal Displacement of B = ? a 0.0223 in. b 0.3064 in. c 0.0333 in. d 0.2304 in. Vertical Displacement of B = ? a 0.2304 in. b 0.0333 in. c 0.076 in. d 0.0555 in. The lower ends of the three bars in Figure SMAT 1810 are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 2 in.², and E = 29 × 10⁶ psi. For the bronze bar, the area is 1.5 in.² and E = 12 × 10⁶ psi. Determine (a) the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar; a 14.97 in. b 1.86 in. c 7.37 in. d 22.34 in. CONT. PREVIOUS PROBLEM: and (b) the length of the bronze that will make the steel stress twice the bronze stress. a 2.48 in. b 3.3 in. c 29.79 in. d 4.15 in. Two plates, shown in Figure STLD 1801 each with thickness t = 16 mm. are bolted together with 6 - 22 mm. diameter bolts forming a lap connection. Bolts are as follows: s1 = 40 mm., s2 = 80 mm. and s3 = 100 mm. Bolt hole diameter = 25 mm. ALLOWABLE STRESSES: Tensile Stress on gross area of the plates = 0.60Fy ; Tensile Stress on net area of the plates = 0.50Fu ; Shear Stress of the bolt (Fv) = 120 MPa ; Bearing Stress of the bolt(Fp) = 1.2Fu. Material used is an A36 steel with Fy = 248 MPa and Fu = 400 MPa. Calculate the permissible tensile load P under the following conditions: (a) Based on shear capacity of bolts.
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a 183.38 kN b 273.7 kN c 90.32 kN d 1013.76 kN (b) Based on the bearing capacity of bolts. a 1013.76 kN b 364.02 kN c 273.7 kN d 457.07 kN (c) Based on block shear stress. a 396.64 kN b 592 kN c 195.36 kN d 768 kN A square hollow steel strut shown in Figure STLD 1802 with a wall thickness, t1 = 10 mm is pin connected to two gusset plates, having a thickness t2 = 12 mm, which are welded to the base plate having a thickness of 12 mm and fastened to a concrete base by 4 - 16 mm diameter anchor bolts. Diameter of pin is 16 mm. Compressive load P = 48 kN, θ = 30°.(a) Calculate the bearing stress between the strut and the pin. a 119.37 MPa. b 158.76 MPa. c 150 MPa. d 199.34 MPa. (b) Calculate the shear stress in the pin. a 51.69 MPa. b 68.74 MPa. c 39.39 MPa. d 119.37 MPa. (c) Calculate the shear stress in the anchor bolt. a 150 MPa. b 199.5 MPa. c 51.69 MPa. d 250.5 MPa. Figure STLD 1803 is an L150×100×6 connected to a larger angular bar. Assume that the tensile force P pass through the centroid such that there is no eccentricity. Determine the safe load P that the member could carry if the allowable tensile stress 148 MPa. Bolt Diameter = 20 mm; Bolt Hole diameter = 22 mm. a 182.26 kN. b 262.19 kN. c 60.15 kN. d 197.14 kN. A staggered connection is shown in Figure STLD 1804. Determine the maximum tensile force that the plate could carry if the allowable tensile stress is 140 MPa. For Plate: t = 12 mm., width = 350 mm. For Rivet: Bolt Diameter = 20 mm. ; Hole diameter = 23 mm. ; Fp = 150 MPa. & Fv = 100 MPa. a 357.03 kN. b 262.19 kN. c 197.14 kN. d 532.89 kN. A tension member made-up of a pair of angles is connected as shown in Figure STLD 1805 with 4 - 25 mm Ø bolts instandard holes. All structural steel is A36. Assuming that the connection between the angles and the structural teeis satisfactory. Allowable bolt shear = 117 MPa.; Allowable tensile stress = 150 MPa.; Allowable bearing stress = 480 MPa. (a) Find the value of P by shear and tension a 287.16 kN. b c 94.76 kN. d (b) Find the value of P by bearing.
300 kN. 501 kN.
a 950 kN. b 1263.5 kN. c 960 kN. d 1586.5 kN. (c) what is the diameter of the bolt if P = 360 kN.? a 18.75 mm. b 30 mm. c 9.24 mm. d 27.99 mm.
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A lapped, bolted tension member is shown in Figure STLD 1806. Diameter of bolts are 18 mm and the plate material is A36 steel, Fy = 250 Mpa. Fu = 400 MPa. Assume the fasteners are adequate and does not contro the tensile capacity. Diameter of hole is 3 mm bigger than the diameter of bolt. (a) Determine the tensile capacity of the lapped joint based on gross area.
a 495000 N. b 526680 N. c 396000 N. d 661320 N. Determine the tensile capacity of the lapped joint based on block shear strength. a 373659 b 570900 c 557700 d 953403 A wood-stave penstock of 1.8-m diameter is wrapped with steel hoops and subjected to a fluid pressure of 380 kPa. If the allowable tension in the hoops is 42.4 kN, what is the maximum hoop spacing in millimeters? a 83.06 mm. b 123.98 mm. c 99.58 mm. d 166.3 mm. The shaft shown in Figure SMAT 1812 is made of bronze and steel and is fixed at both ends. It is subjected to a concentrated torque, 5 kN-m, applied at the junction. The steel segment is 1.5 meters long and the bronze segment is 1.2 meters. For BRONZE: d = 75 mm. ; G = 35 GPa. For STEEL: d = 50 mm. ; G = 83 GPa. (a) Determine the maximum shearing stress in the Steel section. a 55.53 MPa. c 18.33 MPa. (b) Determine the a 16.45 MPa. c 14.49 MPa. (c) Determine the a 1.54 degrees c 0.76 degrees If a force P with position shown in will be 11/3 MPa. respectively. (a) value of x?
b 148.19 MPa. d 247.47 MPa. maximum shearing stress in the bronze section. b 21.88 MPa. d 43.91 MPa. angle of twist in the bronze section. b 0.0401 degrees d 2.3 degrees magnitude equal to 200 kN. is appied at the Figure SMAT 1813, the stress at point A and B compresion and 29/3 MPa. compression, Which of the following most nearly gives the
a 30 mm. b 66.5 mm. c 9.9 mm. d 50 mm. (c) Which of the following most nearly gives the minimum value of the stress?
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a -3 MPa. b -3.99 MPa. c 3 MPa. d -5.01 MPa. Find the maximum deflection for the cantilever beam loaded as shown in Figure TOS 1801 if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa., a = 2 m, b = 1 m., w = 4 kN/m.
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a 18.76 mm. b 0.028 mm. c 28 mm. d 0.0468 mm. For the beam shown in Figure TOS 1802, determine the value of I that will limit the maximum deflection to 0.5 in. Assume that E = 1.5 × 10⁶ psi. w = 60 lb/ft, a = 3 ft, b = 5 ft., c = 2 ft.
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-14.65 in4. -21.86 in4.
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-25.55 in4. -42.66 in4.
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Determine the midspan value of EIδ for the beam shown in Figure TOS 1805. (W = 600 N/m.)
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a -1984.04 N.m3 b -2961.25 N.m3 c 5361.25 N.m3 d 8953.29 N.m3 Determine the value of EIy midway between the supports for the beam loaded as shown in Figure TOS 1806. (M = 1800 N.m., w = 200 N/m.)
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a -1416.67 N.m3 b -1884.17 N.m3 c 764.5 N.m3 d 2316.67 N.m3 If E = 29 × 10⁶ psi, what value of I is required to limit the midspan deflection to 1/360 of the span for the beam in Figure TOS 1807? (w = 2400 lb/ft.)
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a 66.38 in4. b 1059.41 in4. c 796.55 in4. d 1330.24 in4. Find the maximum bending stress in the beam loaded as shown in Figure SMAT 1814. P = 20 N. The beam used is a wide flange section with the following properties: bf = 181 mm; tf = 18.2 mm.; tw = 15.9 mm.; d = 417 mm.
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a 0.0381 MPa. b 0.0331 MPa. c 0.0126 MPa. d 0.0249 MPa. Find the maximum shearing stress in the beam loaded as shown in Figure SMAT 1814. P = 20 N. The beam used is a wide flange section with the following properties: bf = 181 mm; tf = 18.2 mm.; tw = 15.9 mm.; d = 417 mm.
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a 0.008 MPa. b 0.0083 MPa. c 0.0026 MPa. d 0.0062 MPa. A beam with a span of 9 m is fixed at both ends and it carries a uniformly distributed load of W = 25kN/m throughout its length and a superimposed load, 20 kN. at a point 2 meters from the left support. The beam is W 16 × 77 with the following properties: bf = 260 mm.; tf = 19.3 mm.; d = 420 mm.; tw = 11.6 mm.; Ix = 462.02 × 10⁶ mm⁴; Iy = 57.44 × 10⁶ mm⁴. What is the maximum bending stress in the beam? a 58.76 MPa. b 76.7 MPa. c 87.7 MPa. d 128.09 MPa. What is the maximum shearing stress in the beam? a 25.53 MPa. b 28.63 MPa. c 9.45 MPa. d 42.64 MPa. A beam with a cross section shown in Figure STLD 1807 is loaded by a concentrated load, P, at its mid-span. The beam is simply supported over a span of 6 meters. Allowable bending stress are 180 MPa. in tension and 80 MPa. in compression. What is the value of P without exceeding the allowable tensile stress? (Neglect weight of beam.) a 14.2 kN. b 24.8 kN. c 18.65 kN. d 21.19 kN. What is the value of P without exceeding the allowable compressive stress? a 22.46 kN. b 26.28 kN. c 19.76 kN. d 33 kN. Steel and aluminum plates are used to reinforced an 80 mm by 150 mm timber beam. The three materials are fastened firmly as shown in Figure STLD 1808 so that there will be no relative movement between them. [Allowable Bending Stress (Fb) Steel = 120 MPa; Aluminum = 80 MPa; Wood = 10 MPa. Modulus of Elasticity (E) Steel = 200 GPa; Aluminum = 70 GPa; Wood = 10 GPa.] What is the value of the moment of inertia about the neutral axis?
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a 362142444.44 mm⁴. b 671650000 mm⁴. c 505000000 mm⁴. d 540511111.1 mm⁴. What is the safe resisting moment of the beam in kN·m? a 33.41 b 72.25 c 54.32 d 49.86 W 18 × 311 is used as a beam simply supported over a span of 14 meters. The beam caries uniform dead load (including beam weight) of 40kN/m and uniform liveload of 70 kN/m, bothe acting on a plane along its major axis. Allowable bemding stresses Fb = 0.60Fy. Use E = 200 GPa. and Fy = 250 MPa. The properties of the section are as follows: bf = 280 mm.; tf = 75 mm.; d = 620 mm.; tw = 88 mm.; and Ix = 3900×10⁶ mm⁴. The beam is to be reinforced with 42 mm thick plates to be welded on the outer face of each flange. What is the moment capacity of the steel section?
a 1264.29 kN-m. b 2598.82 kN-m. c 1954 kN-m. d 1887 kN-m. A 150 mm by 300 mm wooden beam having a simple span of 6 meters carries a concentrated load P at its midspan. It is notched at the supports. For this problem, all calculations are based on shear alone using the 2010 NSCP specification. Allowable shear stress of wood, Fv = 1.0 MPa. If P = 30 kN, calculate the maximum allowable depth (millimeters) of notches at the supports.
a 15.95 mm. b 23.81 mm. c 19 mm. d 31.73 mm. If the depth of notches is 100 mm, what is the safe value of P (kiloNewton) the beam can carry. a 1.49 kN. b c 2.22 kN. d If P = 25 kN and the depth of notches is the shear stress (MegaPascal) near
1.77 kN. 1.33 kN. is 150 millimeters, what the supports.
a 2.67 MPa. b 3.55 MPa. c 3.33 MPa. d 4.45 MPa. Light-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined 1V:3H and distance between trusses is equal to 3 m. The purlins has a weight of 71 N/m and spaced at 1.2 m on centers. The dead load including for materials is 1200 Pa., live load of 1000 Pa. and wind load of 1440 Pa. Coefficient of pressure at leeward and windward are 0.5 and 0.2 respectively. Sag rods are placed at the middle thirds and Fbx=Fby=138 MPa. (ASSUME ALL FORCES PASSES THROUGH THE CENTROID OF THE SECTION) Sx = 4.48 x 104 mm3 Sy = 1.18 x 104 mm3 Using the interaction formula, determine the following: (a)What is the maximum Ratio of actual to allowable bending stress for combination of (D+L) load?
a 0.5433 b 0.7225 c 0.1632 d 0.4946 (b)What is the maximum Ratio of actual to allowable bending stress for combination of 0.75(D+L+W) load? a 0.4181 b 0.4546 c 0.138 d 0.7592 (c) What is the maximum Ratio of actual to allowable bending stress for combination of (D+L) load if one line of sag rod was placed at mid-span? a c
0.5853 0.1931
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0.7221 1.2059
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A restrained beams with a span of 4 meters has a rectangular cross-section (150 mm. × 250 mm.). The unit weight of the material used is 500 N per cubic meters. The beam is to carry a superimposed uniformly distributed load of 20 N/m. What is the maximum bending stress of on the Beam? a 0.0222 MPa. b 0.0264 MPa. c 0.0109 MPa. d 0.0331 MPa. A restrained beams with a span of 4 meters has a rectangular cross-section (150 mm. × 250 mm.). The unit weight of the material used is 5 N per cubic meters. The beam is to carry a superimposed uniformly-varrying distributed load of 20 N/m from one support an 5 N/m at the other. (a)What is the maximum bending stress of on the Beam? a 0.0187 MPa. b 0.0483 MPa. c 0.0092 MPa. d 0.0607 MPa. (b) What is the moment of inertia with respect to the major axis, Ix? a 130859375 mm4. b 93515625 mm4. c 64453125 mm4. d 117421875 mm4.
Set A
College of Engineering Civil Engineering Department MID PRE-BOARD in Structural Design
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In Figure SMAT 1804, assume that a 25-mm-diameter rivet joins the plates that are each 100 mm wide. The allowable stresses are 150 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; a 7.85 mm. b 17.41 mm. c 13.09 mm. d 21.86 mm. and (b) the largest average tensile stress in the plates. a 18 Mpa. b 50 Mpa. c 16.5 Mpa. d 30.06 Mpa. Given the Figure SMAT 1809: Vertical load P = 6000 lb; Crosssectional area of each rod = 0.5 in²; E = 10 × 10⁶ psi; α = 30°; θ = 30°; Required: (a)Deformation ∆ AB = ? a 0.144 in. b -0.1149 in. c -0.0864 in. d -0.1443 in. Deformation ∆ BC = ? a 0.144 in. b -0.0864 in. c -0.0285 in. d 0.2405 in. Horizontal Displacement of B = ? a 0.0223 in. b 0.3064 in. c 0.0333 in. d 0.2304 in. Vertical Displacement of B = ? a 0.2304 in. b 0.0333 in. c 0.076 in. d 0.0555 in. The lower ends of the three bars in Figure SMAT 1810 are at the same level before the uniform rigid block weighing 40 kips is attached. Each steel bar has a length of 3 ft, and area of 2 in.², and E = 29 × 10⁶ psi. For the bronze bar, the area is 1.5 in.² and E = 12 × 10⁶ psi. Determine (a) the length of the bronze bar so that the load on each steel bar is twice the load on the bronze bar; a 14.97 in. b 1.86 in. c 7.37 in. d 22.34 in. CONT. PREVIOUS PROBLEM: and (b) the length of the bronze that will make the steel stress twice the bronze stress. a 2.48 in. b 3.3 in. c 29.79 in. d 4.15 in. Two plates, shown in Figure STLD 1801 each with thickness t = 16 mm. are bolted together with 6 - 22 mm. diameter bolts forming a lap connection. Bolts are as follows: s1 = 40 mm., s2 = 80 mm. and s3 = 100 mm. Bolt hole diameter = 25 mm. ALLOWABLE STRESSES: Tensile Stress on gross area of the plates = 0.60Fy ; Tensile Stress on net area of the plates = 0.50Fu ; Shear Stress of the bolt (Fv) = 120 MPa ; Bearing Stress of the bolt(Fp) = 1.2Fu. Material used is an A36 steel with Fy = 248 MPa and Fu = 400 MPa. Calculate the permissible tensile load P under the following conditions: (a) Based on shear capacity of bolts.
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a 183.38 kN b 273.7 kN c 90.32 kN d 1013.76 kN (b) Based on the bearing capacity of bolts. a 1013.76 kN b 364.02 kN c 273.7 kN d 457.07 kN (c) Based on block shear stress. a 396.64 kN b 592 kN c 195.36 kN d 768 kN A square hollow steel strut shown in Figure STLD 1802 with a wall thickness, t1 = 10 mm is pin connected to two gusset plates, having a thickness t2 = 12 mm, which are welded to the base plate having a thickness of 12 mm and fastened to a concrete base by 4 - 16 mm diameter anchor bolts. Diameter of pin is 16 mm. Compressive load P = 48 kN, θ = 30°.(a) Calculate the bearing stress between the strut and the pin. a 119.37 MPa. b 158.76 MPa. c 150 MPa. d 199.34 MPa. (b) Calculate the shear stress in the pin. a 51.69 MPa. b 68.74 MPa. c 39.39 MPa. d 119.37 MPa. (c) Calculate the shear stress in the anchor bolt. a 150 MPa. b 199.5 MPa. c 51.69 MPa. d 250.5 MPa. Figure STLD 1803 is an L150×100×6 connected to a larger angular bar. Assume that the tensile force P pass through the centroid such that there is no eccentricity. Determine the safe load P that the member could carry if the allowable tensile stress 148 MPa. Bolt Diameter = 20 mm; Bolt Hole diameter = 22 mm. a 182.26 kN. b 262.19 kN. c 60.15 kN. d 197.14 kN. A staggered connection is shown in Figure STLD 1804. Determine the maximum tensile force that the plate could carry if the allowable tensile stress is 140 MPa. For Plate: t = 12 mm., width = 350 mm. For Rivet: Bolt Diameter = 20 mm. ; Hole diameter = 23 mm. ; Fp = 150 MPa. & Fv = 100 MPa. a 357.03 kN. b 262.19 kN. c 197.14 kN. d 532.89 kN. A tension member made-up of a pair of angles is connected as shown in Figure STLD 1805 with 4 - 25 mm Ø bolts instandard holes. All structural steel is A36. Assuming that the connection between the angles and the structural teeis satisfactory. Allowable bolt shear = 117 MPa.; Allowable tensile stress = 150 MPa.; Allowable bearing stress = 480 MPa. (a) Find the value of P by shear and tension a 287.16 kN. b c 94.76 kN. d (b) Find the value of P by bearing.
300 kN. 501 kN.
a 950 kN. b 1263.5 kN. c 960 kN. d 1586.5 kN. (c) what is the diameter of the bolt if P = 360 kN.? a 18.75 mm. b 30 mm. c 9.24 mm. d 27.99 mm.
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A lapped, bolted tension member is shown in Figure STLD 1806. Diameter of bolts are 18 mm and the plate material is A36 steel, Fy = 250 Mpa. Fu = 400 MPa. Assume the fasteners are adequate and does not contro the tensile capacity. Diameter of hole is 3 mm bigger than the diameter of bolt. (a) Determine the tensile capacity of the lapped joint based on gross area.
a 495000 N. b 526680 N. c 396000 N. d 661320 N. Determine the tensile capacity of the lapped joint based on block shear strength. a 373659 b 570900 c 557700 d 953403 A wood-stave penstock of 1.8-m diameter is wrapped with steel hoops and subjected to a fluid pressure of 380 kPa. If the allowable tension in the hoops is 42.4 kN, what is the maximum hoop spacing in millimeters? a 83.06 mm. b 123.98 mm. c 99.58 mm. d 166.3 mm. The shaft shown in Figure SMAT 1812 is made of bronze and steel and is fixed at both ends. It is subjected to a concentrated torque, 5 kN-m, applied at the junction. The steel segment is 1.5 meters long and the bronze segment is 1.2 meters. For BRONZE: d = 75 mm. ; G = 35 GPa. For STEEL: d = 50 mm. ; G = 83 GPa. (a) Determine the maximum shearing stress in the Steel section. a 55.53 MPa. c 18.33 MPa. (b) Determine the a 16.45 MPa. c 14.49 MPa. (c) Determine the a 1.54 degrees c 0.76 degrees If a force P with position shown in will be 11/3 MPa. respectively. (a) value of x?
b 148.19 MPa. d 247.47 MPa. maximum shearing stress in the bronze section. b 21.88 MPa. d 43.91 MPa. angle of twist in the bronze section. b 0.0401 degrees d 2.3 degrees magnitude equal to 200 kN. is appied at the Figure SMAT 1813, the stress at point A and B compresion and 29/3 MPa. compression, Which of the following most nearly gives the
a 30 mm. b 66.5 mm. c 9.9 mm. d 50 mm. (c) Which of the following most nearly gives the minimum value of the stress?
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a -3 MPa. b -3.99 MPa. c 3 MPa. d -5.01 MPa. Find the maximum deflection for the cantilever beam loaded as shown in Figure TOS 1801 if the cross section is 50 mm wide by 150 mm high. Use E = 69 GPa., a = 2 m, b = 1 m., w = 4 kN/m.
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a 18.76 mm. b 0.028 mm. c 28 mm. d 0.0468 mm. For the beam shown in Figure TOS 1802, determine the value of I that will limit the maximum deflection to 0.5 in. Assume that E = 1.5 × 10⁶ psi. w = 60 lb/ft, a = 3 ft, b = 5 ft., c = 2 ft.
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-14.65 in4. -21.86 in4.
b d
-25.55 in4. -42.66 in4.
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Determine the midspan value of EIδ for the beam shown in Figure TOS 1805. (W = 600 N/m.)
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a -1984.04 N.m3 b -2961.25 N.m3 c 5361.25 N.m3 d 8953.29 N.m3 Determine the value of EIy midway between the supports for the beam loaded as shown in Figure TOS 1806. (M = 1800 N.m., w = 200 N/m.)
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a -1416.67 N.m3 b -1884.17 N.m3 c 764.5 N.m3 d 2316.67 N.m3 If E = 29 × 10⁶ psi, what value of I is required to limit the midspan deflection to 1/360 of the span for the beam in Figure TOS 1807? (w = 2400 lb/ft.)
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a 66.38 in4. b 1059.41 in4. c 796.55 in4. d 1330.24 in4. Find the maximum bending stress in the beam loaded as shown in Figure SMAT 1814. P = 20 N. The beam used is a wide flange section with the following properties: bf = 181 mm; tf = 18.2 mm.; tw = 15.9 mm.; d = 417 mm.
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a 0.0381 MPa. b 0.0331 MPa. c 0.0126 MPa. d 0.0249 MPa. Find the maximum shearing stress in the beam loaded as shown in Figure SMAT 1814. P = 20 N. The beam used is a wide flange section with the following properties: bf = 181 mm; tf = 18.2 mm.; tw = 15.9 mm.; d = 417 mm.
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a 0.008 MPa. b 0.0083 MPa. c 0.0026 MPa. d 0.0062 MPa. A beam with a span of 9 m is fixed at both ends and it carries a uniformly distributed load of W = 25kN/m throughout its length and a superimposed load, 20 kN. at a point 2 meters from the left support. The beam is W 16 × 77 with the following properties: bf = 260 mm.; tf = 19.3 mm.; d = 420 mm.; tw = 11.6 mm.; Ix = 462.02 × 10⁶ mm⁴; Iy = 57.44 × 10⁶ mm⁴. What is the maximum bending stress in the beam? a 58.76 MPa. b 76.7 MPa. c 87.7 MPa. d 128.09 MPa. What is the maximum shearing stress in the beam? a 25.53 MPa. b 28.63 MPa. c 9.45 MPa. d 42.64 MPa. A beam with a cross section shown in Figure STLD 1807 is loaded by a concentrated load, P, at its mid-span. The beam is simply supported over a span of 6 meters. Allowable bending stress are 180 MPa. in tension and 80 MPa. in compression. What is the value of P without exceeding the allowable tensile stress? (Neglect weight of beam.) a 14.2 kN. b 24.8 kN. c 18.65 kN. d 21.19 kN. What is the value of P without exceeding the allowable compressive stress? a 22.46 kN. b 26.28 kN. c 19.76 kN. d 33 kN. Steel and aluminum plates are used to reinforced an 80 mm by 150 mm timber beam. The three materials are fastened firmly as shown in Figure STLD 1808 so that there will be no relative movement between them. [Allowable Bending Stress (Fb) Steel = 120 MPa; Aluminum = 80 MPa; Wood = 10 MPa. Modulus of Elasticity (E) Steel = 200 GPa; Aluminum = 70 GPa; Wood = 10 GPa.] What is the value of the moment of inertia about the neutral axis?
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a 362142444.44 mm⁴. b 671650000 mm⁴. c 505000000 mm⁴. d 540511111.1 mm⁴. What is the safe resisting moment of the beam in kN·m? a 33.41 b 72.25 c 54.32 d 49.86 W 18 × 311 is used as a beam simply supported over a span of 14 meters. The beam caries uniform dead load (including beam weight) of 40kN/m and uniform liveload of 70 kN/m, bothe acting on a plane along its major axis. Allowable bemding stresses Fb = 0.60Fy. Use E = 200 GPa. and Fy = 250 MPa. The properties of the section are as follows: bf = 280 mm.; tf = 75 mm.; d = 620 mm.; tw = 88 mm.; and Ix = 3900×10⁶ mm⁴. The beam is to be reinforced with 42 mm thick plates to be welded on the outer face of each flange. What is the moment capacity of the steel section?
a 1264.29 kN-m. b 2598.82 kN-m. c 1954 kN-m. d 1887 kN-m. A 150 mm by 300 mm wooden beam having a simple span of 6 meters carries a concentrated load P at its midspan. It is notched at the supports. For this problem, all calculations are based on shear alone using the 2010 NSCP specification. Allowable shear stress of wood, Fv = 1.0 MPa. If P = 30 kN, calculate the maximum allowable depth (millimeters) of notches at the supports.
a 15.95 mm. b 23.81 mm. c 19 mm. d 31.73 mm. If the depth of notches is 100 mm, what is the safe value of P (kiloNewton) the beam can carry. a 1.49 kN. b c 2.22 kN. d If P = 25 kN and the depth of notches is the shear stress (MegaPascal) near
1.77 kN. 1.33 kN. is 150 millimeters, what the supports.
a 2.67 MPa. b 3.55 MPa. c 3.33 MPa. d 4.45 MPa. Light-grade steel channel was used as a purlin of a truss. The top chord of the truss is inclined 1V:3H and distance between trusses is equal to 3 m. The purlins has a weight of 71 N/m and spaced at 1.2 m on centers. The dead load including for materials is 1200 Pa., live load of 1000 Pa. and wind load of 1440 Pa. Coefficient of pressure at leeward and windward are 0.5 and 0.2 respectively. Sag rods are placed at the middle thirds and Fbx=Fby=138 MPa. (ASSUME ALL FORCES PASSES THROUGH THE CENTROID OF THE SECTION) Sx = 4.48 x 104 mm3 Sy = 1.18 x 104 mm3 Using the interaction formula, determine the following: (a)What is the maximum Ratio of actual to allowable bending stress for combination of (D+L) load?
a 0.5433 b 0.7225 c 0.1632 d 0.4946 (b)What is the maximum Ratio of actual to allowable bending stress for combination of 0.75(D+L+W) load? a 0.4181 b 0.4546 c 0.138 d 0.7592 (c) What is the maximum Ratio of actual to allowable bending stress for combination of (D+L) load if one line of sag rod was placed at mid-span? a c
0.5853 0.1931
b d
0.7221 1.2059
48 .
49 .
50 .
A restrained beams with a span of 4 meters has a rectangular cross-section (150 mm. × 250 mm.). The unit weight of the material used is 500 N per cubic meters. The beam is to carry a superimposed uniformly distributed load of 20 N/m. What is the maximum bending stress of on the Beam? a 0.0222 MPa. b 0.0264 MPa. c 0.0109 MPa. d 0.0331 MPa. A restrained beams with a span of 4 meters has a rectangular cross-section (150 mm. × 250 mm.). The unit weight of the material used is 5 N per cubic meters. The beam is to carry a superimposed uniformly-varrying distributed load of 20 N/m from one support an 5 N/m at the other. (a)What is the maximum bending stress of on the Beam? a 0.0187 MPa. b 0.0483 MPa. c 0.0092 MPa. d 0.0607 MPa. (b) What is the moment of inertia with respect to the major axis, Ix? a 130859375 mm4. b 93515625 mm4. c 64453125 mm4. d 117421875 mm4.
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