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Spiral reinforced columns are more ductile than tied columns, allowing them a _______ strength-reduction factor.

A wall footing has the following conditions. Determine the critical one-way shear, Vu, at a distance of d from the face of the wall. Assume the footing is 4 ft wide, the pressure that acts on the bottom of the footing is 5,900 psf, and the reinforcement is a #6 bar spaced every 12 inches.The bottom of the footing is at a depth of 4 ft below grade.The service dead load is 9 kips/ft, and the service live load is 8 kips/ft.The wall is 16 in. thick.The footing is 10 in. thick.The allowable soil pressure, qa, is 5,000 psf.The soil has a density of 120 lb/ft3.The concrete has a density of 150 lb/ft3.The concrete cover has a thickness of 3 in.f’c = 3,500 psi and fy = 60,000 psi.

Tied columns tend to hold up better during seismic events compared to traditional spiral columns.

To account for the effect of accidental moments, the ACI Code specifies that the maximum load on a column must not exceed 0.85 times the axial-load capacity of spiral columns and 0.80 times the axial-load capacity of tied columns.

A wall footing has the following conditions. Determine the minimum cross-sectional area of the reinforcement. Assume the footing is 6 ft wide and the pressure that acts on the bottom of the footing is 4,133 psf.The bottom of the footing is at a depth of 5 ft below grade.The service dead load is 10 kips/ft, and the service live load is 8 kips/ft.The wall is 14 in. thick.The footing is 10 in. thick.The allowable soil pressure, qa, is 4,100 psf.The soil has a density of 115 lb/ft3.The concrete has a density of 150 lb/ft3.The concrete cover has a thickness of 3 in.f’c = 3,100 psi and fy = 60,000 psi.

A wall footing has the following conditions. Determine the critical one-way shear, Vu, at a distance of d from the face of the wall. Assume the footing is 4 ft wide, the pressure that acts on the bottom of the footing is 4,599 psf, and the reinforcement is a #4 bar spaced every 12 inches.The bottom of the footing is at a depth of 3 ft below grade.The service dead load is 6 kips/ft, and the service live load is 7 kips/ft.The wall is 14 in. thick.The footing is 10 in. thick.The allowable soil pressure, qa, is 4,400 psf.The soil has a density of 125 lb/ft3.The concrete has a density of 150 lb/ft3.The concrete cover has a thickness of 3 in.f’c = 3,800 psi and fy = 60,000 psi.

A wall footing has the following conditions. Determine the moment capacity, Mn, of the footing. Assume the footing is 4 ft wide, the pressure that acts on the bottom of the footing is 6,500 psf, and the reinforcement is a #4 bar spaced every 12 inches.The bottom of the footing is at a depth of 4 ft below grade.The service dead load is 11 kips/ft, and the service live load is 8 kips/ft.The wall is 10 in. thick.The footing is 13 in. thick.The allowable soil pressure, qa, is 5,800 psf.The soil has a density of 120 lb/ft3.The concrete has a density of 150 lb/ft3.The concrete cover has a thickness of 3 in.f’c = 3,800 psi and fy = 60,000 psi.

Solving a design problem typically involves guessing a section, analyzing whether it will be satisfactory, revising the section, reanalyzing it.

Determine the required splice length for the galvanized longitudinal bars in the following tied column made of normal-weight concrete. Assume that some of the longitudinal bars are in tension and that the connection should be designed as a Class B contact lap splice. Use ACI 318-14 Table 25.4.2.2 to calculate the development length, and assume that the ties satisfy the Code minimum.Column width, b = 20 in.Column thickness, h = 28 in.Clear cover to ties = 1.5 in.Number of longitudinal bars = 10Size of longitudinal bars = No. 10Size of ties = No. 3Concrete strength = 8,500 psiYield strength of longitudinal bars = 60,000 psiYield strength of ties = 40,000 psi

Determine the required splice length for the uncoated longitudinal bars in the following tied column made of normal-weight concrete. Assume that some of the longitudinal bars are in tension and that the connection should be designed as a Class B contact lap splice. Use ACI 318-14 Table 25.4.2.2 to calculate the development length, and assume that the ties satisfy the Code minimum.Column width, b = 20 in.Column thickness, h = 18 in.Clear cover to ties = 1.5 in.Number of longitudinal bars = 10Size of longitudinal bars = No. 7Size of ties = No. 4Concrete strength = 4,500 psiYield strength of longitudinal bars = 60,000 psiYield strength of ties = 40,000 psi