A simply supported beam is subjected to uniformly distributed loads: a service dead load [tex]w_D=5 kips / ft[/tex] and a service live load [tex]w_L=7 kips / ft[/tex]. Use the load combination [tex]w_u=1.2 w_D+ 1.6 w_L[/tex]. Note that the shear force is expressed as [tex]V_u=w_u(\frac{L}{2}-x)[/tex] where [tex]x[/tex] is the distance from a support. (65pts total)
The beam also has the following properties.
[tex]
\begin{array}{c}
f_c=4 ksi, \quad f_y=f_{yt}=60 ksi, \quad b=18 in \
h=26 in, \quad d=24.5 in
\end{array}
[/tex]
No. 4 bars, U-stirrups used, Normal weight concrete
(1) Calculate the required spacing of stirrups. (15pts)
Answer (2)
The required spacing of stirrups for the simply supported beam, based on calculated loads and shear strengths, is 12.25 inches. This is determined by using the factored uniform load and the properties of the materials involved. Since minimum spacing is mandated by code, and it does not exceed the calculated maximum, 12.25 inches is both effective and within regulatory requirements.
;
Calculate the factored uniform load: w u = 1.2 w D + 1.6 w L = 17.2 ki p s / f t .
Calculate the concrete shear strength: V c = 2 λ f c b d = 1764 ki p s .
Calculate the shear force at distance d: V u = w u ( L /2 − d ) = 222.88 ki p s (assuming L=30 ft).
Since V u < ϕ V c , minimum stirrups are required, and the maximum spacing is determined by code requirements, resulting in a spacing of 12.25 in .
Explanation
Problem Setup We are given a simply supported beam with a uniformly distributed dead load w D = 5 ki p s / f t and a live load w L = 7 ki p s / f t . We need to calculate the required spacing of stirrups using the load combination w u = 1.2 w D + 1.6 w L . We are also given the concrete strength f c = 4 k s i , steel yield strength f y = f y t = 60 k s i , beam width b = 18 in , and effective depth d = 24.5 in . The stirrups are No. 4 U-stirrups.
Calculate Factored Load First, calculate the factored uniform load: w u = 1.2 w D + 1.6 w L = 1.2 ( 5 ki p s / f t ) + 1.6 ( 7 ki p s / f t ) = 6 + 11.2 = 17.2 ki p s / f t
Calculate Concrete Shear Strength Next, calculate the concrete shear strength: V c = 2 λ f c b d
Since we have normal weight concrete, λ = 1.0 . Therefore, V c = 2 ( 1.0 ) 4 ( 18 in ) ( 24.5 in ) = 2 ( 2 ) ( 18 ) ( 24.5 ) = 1764 ki p s
Calculate Shear Force at Distance d Now, we need to determine the shear force V u at a distance d from the support. We are given that V u = w u ( L /2 − x ) , where x is the distance from the support. We will assume a reasonable beam length of L = 30 f t . Then, at x = d = 24.5 in = 2.04 f t ,
V u = w u ( L /2 − d ) = 17.2 ki p s / f t ( 30 f t /2 − 2.04 f t ) = 17.2 ( 15 − 2.04 ) = 17.2 ( 12.96 ) = 222.88 ki p s
Calculate Design Shear Strength Calculate the design shear strength ϕ V c , where ϕ = 0.75 for shear: ϕ V c = 0.75 ( 1764 ki p s ) = 1323 ki p s
Calculate Required Shear Strength by Stirrups Now, we calculate the shear strength that needs to be provided by the stirrups, V s . Since V u < ϕ V c , stirrups are not theoretically required by code. However, minimum stirrups are still required. We proceed by calculating V s :
V s = ϕ V u − ϕ V c = 0.75 222.88 − 1323 = 0.75 − 1100.12 = − 1466.83 ki p s
Since V s is negative, this confirms that stirrups are not required for shear strength, but minimum stirrups are still needed.
Determine Maximum Stirrup Spacing For No. 4 U-stirrups, the area of steel is A v = 2 ∗ 0.20 i n 2 = 0.40 i n 2 . We need to find the maximum stirrup spacing. First, we calculate the maximum spacing allowed by code: s ma x = min ( 2 d , 24 in ) = min ( 2 24.5 , 24 ) = min ( 12.25 , 24 ) = 12.25 in Also, we need to check if V s ≤ 4 f c b d = 4 4 ( 18 ) ( 24.5 ) = 4 ( 2 ) ( 18 ) ( 24.5 ) = 3528 ki p s . Since V s = − 1466.83 ki p s < 3528 ki p s , we use s ma x = 12.25 in .
Calculate Minimum Stirrup Spacing Since V u < ϕ V c , we provide minimum stirrups. The ACI code specifies the following requirement for minimum stirrups: A v , min = 0.75 f c f y t b w s ≥ 50 f y t b w s Solving for s: s ≤ 0.75 f c b w A v f y t = 0.75 4000 ∗ 18 0.4 ∗ 60000 = 0.75 ∗ 20 ∗ 18 24000 = 270 24000 = 88.89 in s ≤ 50 b w A v f y t = 50 ∗ 18 0.4 ∗ 60000 = 900 24000 = 26.67 in Therefore, the maximum spacing for minimum stirrups is the smallest of these values and s ma x = 12.25 in . So, s = min ( 88.89 , 26.67 , 12.25 ) = 12.25 in .
Final Answer Therefore, the required spacing of the stirrups is 12.25 inches.
Examples
Imagine you're designing the support structure for a bridge. Calculating stirrup spacing in the concrete beams, just like in this problem, ensures the bridge can withstand the expected loads from traffic and environmental factors. By properly spacing the stirrups, you prevent shear failure and ensure the bridge's long-term stability and safety. This careful design approach is crucial for maintaining public safety and infrastructure integrity.
