Reinforced Concrete Design Flashcards
6 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 Reinforced Concrete Design flashcards as text
According to ACI 318, what is the primary purpose of providing minimum flexural reinforcement (As,min) in a reinforced concrete beam?
Answer: To prevent a sudden, brittle failure immediately after the concrete cracks.
ACI 318 requires minimum flexural reinforcement to ensure that the nominal flexural strength (Mn) of the cracked section is greater than the cracking moment (Mcr). This ensures that once the concrete's tensile strength is exceeded and it cracks, the reinforcement is sufficient to carry the tensile force, leading to a ductile failure mode characterized by yielding of the steel, rather than a sudden, brittle failure of the unreinforced section.
A rectangular concrete beam has a width (bw) of 12 inches, an effective depth (d) of 20 inches, and is made with 4,000 psi concrete and Grade 60 reinforcement. At a critical section, the factored shear force (Vu) is 52 kips. The design shear strength provided by the concrete (φVc) is 30 kips. Using No. 3 vertical U-stirrups, what is the required spacing (s) to resist the factored shear?
Answer: 9 in
First, determine the shear strength required from the stirrups (φVs). The total design shear strength (φVn) must be greater than or equal to the factored shear (Vu). So, Vu ≤ φVc + φVs. Rearranging gives: φVs ≥ Vu - φVc = 52 kips - 30 kips = 22 kips. The area of a No. 3 U-stirrup (two legs) is Av = 2 * 0.11 in² = 0.22 in². The formula for stirrup spacing is s = (φ * Av * fyt * d) / φVs. Using φ=0.75 for shear, s = (0.75 * 0.22 in² * 60 ksi * 20 in) / 22 kips = 198 / 22 = 9.0 inches. A spacing of 9 inches is required.
Which of the following modifications would result in the SHORTEST required development length for a No. 8 reinforcing bar in tension, assuming all other factors are held constant?
Answer: Using uncoated (black) steel instead of epoxy-coated steel.
The development length equation includes several modification factors. Epoxy coating reduces the bond and increases the required length (ψe > 1.0). The 'top bar' effect also reduces bond and increases the required length (ψt > 1.0). Lightweight aggregate concrete has lower bond strength and increases the required length (λ > 1.0). Decreasing f'c increases the required length. Using uncoated (black) steel provides a better bond (ψe = 1.0) compared to epoxy-coated steel, thus resulting in the shortest development length among the choices.
An engineer is checking the adequacy of a tied reinforced concrete column using its design interaction diagram. The column is subjected to a factored axial load (Pu) of 400 kips and a factored moment (Mu) of 200 kip-ft. The point representing (Mu, Pu) on the diagram plots inside the design strength curve (φMn, φPn). What does this indicate?
Answer: The column design is adequate to resist the applied loads.
A column interaction diagram represents the boundary of all possible combinations of design axial strength (φPn) and design moment strength (φMn). If the point representing the required strength (Pu, Mu) from the factored loads falls inside or on this boundary, it means the column's design strength is greater than the required strength, and the design is therefore adequate and safe. Points outside the curve represent failure.
According to ACI 318, for a solid, non-prestressed one-way slab not supporting or attached to partitions liable to be damaged by large deflections, what is the minimum required thickness (h) for a simply supported span of length L, assuming normal weight concrete and Grade 60 reinforcement?
Answer: L/20
ACI 318 Table 7.3.1.1 provides minimum thickness values for non-prestressed beams and one-way slabs as a simplified method to control deflections without performing detailed calculations. For a simply supported one-way slab using Grade 60 reinforcement, the minimum required thickness (h) is the span length (L) divided by 20. The other values correspond to different support conditions: L/24 is for one end continuous, L/28 is for both ends continuous, and L/16 is for a cantilever.
Which of the following BEST describes the primary purpose of the strength reduction factor (φ) in the LRFD design of reinforced concrete members?
Answer: To account for potential under-strength members due to variations in material strengths, dimensions, and inaccuracies in design equations.
In Load and Resistance Factor Design (LRFD), uncertainties are handled on both sides of the design equation (φRn ≥ U). Load factors (e.g., 1.2D + 1.6L) are applied to service loads to account for uncertainties in loading. The strength reduction factor (φ) is applied to the nominal resistance (Rn) to account for uncertainties on the member strength side, such as materials being weaker than specified (f'c, fy), member dimensions being slightly off from the plans, and approximations made in the design formulas.