steel exam

7/29/2019 Steel exam

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TEST I

Problem # 1A plate is used as a bracket and is attached to a column flange as shown in the figure 1. Bolts

are 30 mm in diameter is 16 mm thick. All distance in mm.

1.

Calculate the minimum bolt shear force.2. Calculate the maximum bolt shear stress.3. Calculate the maximum plate bearing stress.Problem # 2

The composite section shown in figure 2 made up of C 15x20 and W 21x83.Determine

the properties of the composite section.

Problem # 3Assuming that the joint shown will not fail by tearing diagonally from rivet to rivet or

by tearing out the plate behind the rivets, and using the following data, determine the

allowable load and factor and factor of safety of the butt joint.Width of plate = 225mm, thickness of main plate = 12mm, of cover plates = 6mm,

diameter of rivets = 20mm, allowable ft = 110.3MPa, fb = 138MPa, and fs = 69MPa.Ultimate strength: tension = 414MPa bearing = 621MPa, and shearing = 345MPa. Assume

rivet holes to be 3mm larger than the rivets. Refer to figure 3.

Problem # 4

Design the most economical W200 or W250 shape to support a tension of 450kN and 6m

long. There are two rows of 20-mm bolts on each side of the flange. For steel, F y = 250MPaand F u = 400MPa.

Problem # 5The 76mm x 76mm x 6mm single angle shown figure 5 is welded to an 8-mm-thick gusset

plate. Steel is A36 with F y = 248MPa and F u = 400MPa. The electrode used is E60 with F u= 414MPa. Angle cross-sectional area is 926 mm

2.

1. Determine the value of P based on gross area.2. Determine the value of P based on net area using a strength reduction coefficient.3. Determine the value of P based on block shear in gusset plate along the weld.

Problem # 6Steel tire, 12 mm thick, 110 mm wide and of 800 mm inside diameter, is heated and

shrunk onto a steel wheel 800.5 mm in diameter. Neglect the deformation of the wheel. Modulusof elasticity of steel is 200GPa.

1. What is the tensile stress in the tire, in MPa?2. What is the compressive pressure between the tire and the wheel, in MPa?3. If the allowable tensile stress on steel tire is 124MPa, determine its thickness to resist a

pressure of 1.5MPa.


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TEST II. Identification

1. A structural member whose primary function is to carry loads transverse to itslongitudinal.

2. A structural member whose primary function is to carry loads both transverse and parallelto its longitudinal axis.

3. A structural member whose primary function is to carry loads parallel its longitudinal.4. Determination of load effects (force, moment, stress as an appropriate) on members and

connection based on the assumption that material deformation disappears on removal of

force that produce it.

5. Generic term for welds, bolts, rivets or other connecting device.6. Forces or other actions arise on structural systems from the weight of all permanent

construction, occupants and their possessions, environmental effects, differential

settlement and restrained dimensional changes. Permanent loads are in which variation in

time are rare of small magnitude. All other loads are variable loads.

7. A method of proportioning structural components (members, connectors, connectingelements and assemblages) such that no applicable limit state is exceeded when the

structure is subjected to all appropriate load combinations.

8. The capacity of a structure or a component to resist the effects if loads, as determined bythe computations using specified materials strength and dimensions and formulas derived

from the accepted principles of structural mechanics or by field tests or laboratory tests

off scaled models, allowing for modelling effects and differences between laboratory and

field conditions.

9. A wall mat in its own plane resists shear forces resulting from applied wind, earthquakeor other transverse loads or provides frame stability. Also called structural wall.

10.The resistance to deformation of a member or structured measured by the ratio of theapplied force to the corresponding displacement.


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Figure 1 Figure 2

Figure 3 Figure 5

steel exam

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