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1.49 Segment AB of the bar is a tube with an outer diameter of 1.5 in. and a wall thickness of 0.125 in. Segment BC is a solid rod of diameter 0.75 in. Determine the normal stress in each segment.

Given:

Required:

Outside diameter of segment AB (D) = 1.5 in.

Normal stress of AB (τAB) = ?

Thickness of segment AB (t) = 0.125 in.

Normal stress of BC (τBC) = ?

Diameter of segment BC = 0.75 in.

FBD:

Solution: PAB = 4800 – 2800 = 2000lb ( C ) PBC = 2800lb ( T ) σAB =

AB

=

σBC =

BC

=

= 3700 psi (C ) ans. = 6340 psi ( T )

ans.

1.50 The cylindrical steel column has an outer diameter of 4 in. and inner diameterof 3.5 in. The column is separated from the concrete foundation by a square bearing plate. The working compressive stress is 26 000 psi for the column, and the working bearing stress is 1200 psi for concrete. Find the largest force P that can be applied to the column.

Given:

Required : Largest force P

Outside diameter : 4.0 inches Inner diameter : 3.5 inches Compressive stress : 26 000 psi Bearing Stress : 1200 psi

Solution: Assuming that stress in steel column governs: P = σstAst = (26000) P = 76600 lb Assuming that bearing stress on concrete governs: P = σbApl = 1200(72) = 58800 lb

ans.

1.51 The tubular tension member is fabricated by welding a steel strip into a 12_ helix. The cross-sectional area of the resulting tube is 2.75 in.2. If the normal stress acting on the plane of the weld is 12 ksi, determine (a) the axial force P; and (b) the shear stress acting on the plane of the weld.

Given : Angle : 12 degrees Area : 2.75 in. squared Normal stress : 12ksi Required : axial force , shear stress Solution: A= (a) σ = N/A

= 2.811 in.2 (included cross-sectional area)

12 = P = 34.49 kips

ans.

(b) = V/A =

=

= 2.55 kal

ans.

1.52 An aluminum cable of 6 mm diameter is suspended from a high-altitude balloon. The density of aluminum is 2700 kg/m3, and its breaking stress is 390 MPa. Determine the largest length of cable that can be suspended without breaking. Given : Diameter : 6mm Density : 2700 kg/m3 Breaking stress : 390 MPa Required: Largest length Solution: Maximum acial force equals weight of cable: Pmax = pgAL σmax = Pmax/A = pgL 390 x 106 = 2700(9.81)L L = 14720 m = 14.72km Result is independent of diameter of cable

ans.