Stirrup Spacing Calculation For Rectangular Beam Under Shear And Torsion

Hey guys! Today, we're diving into a classic structural engineering problem: figuring out the stirrup spacing for a rectangular beam that's dealing with both shear forces and torsional moments. This is super important in ensuring the beam's safety and preventing catastrophic failures. We'll be walking through the calculations step-by-step, using the ACI code as our guide. So, buckle up, and let's get started!

In this article, we'll tackle a practical problem involving a rectangular beam subjected to factored shear and torsional moment. The goal is to determine the appropriate spacing for closed stirrups to ensure the beam's structural integrity under these combined loads. We'll use industry-standard formulas and design principles, making it easy to follow along and apply to your own projects. This is essential knowledge for civil engineers and anyone involved in structural design. Understanding these principles is critical for designing safe and reliable concrete structures.

Alright, let's lay out the problem. We have a rectangular beam with a width (${b_w}$) of 15 inches and a depth (${h}$) of 30 inches. The concrete compressive strength ( ${f_c^{\prime}}$) is 4000 psi, and the steel yield strength (${f_{yz}}$) is 60,000 psi. The beam is subjected to a factored shear (${V_u}$) of 80 kips and a factored torsional moment (${T_u}$) of 900 in-kips. We need to figure out the required spacing for No. 4 (No. 13) closed stirrups. No. 4 stirrups have a diameter of 0.5 inches, which means their cross-sectional area (${A_v}$) is 0.20 square inches for two legs. This is crucial information for our calculations.

The first step in our calculations is to determine the nominal shear strength (${V_n}$) that the concrete and stirrups need to provide. To do this, we'll use the following formula derived from ACI code provisions:

${ V_n = \frac{V_u}{\phi} }$

Where:

• ${V_u}$ is the factored shear force (80 kips).
• ${\phi}$ is the shear strength reduction factor, which is typically 0.75 for shear and torsion in reinforced concrete design, as per ACI code. This factor accounts for uncertainties in material strengths and construction practices.

Plugging in the values, we get:

${ V_n = \frac{80 \text{ kips}}{0.75} = 106.67 \text{ kips} }$

So, the nominal shear strength (${V_n}$) required is 106.67 kips. This value represents the total shear resistance that the beam must provide, considering both the concrete's contribution and the stirrups' contribution. It's a critical benchmark for our design.

Now, let's figure out how much shear the concrete itself can handle. We'll use the ACI code equation for the concrete shear capacity (${V_c}$):

${ V_c = 2\lambda \sqrt{f_c^{\prime}} b_w d }$

Where:

• ${\lambda}$ is the modification factor for lightweight concrete (1.0 for normal weight concrete).
• ${f_c^{\prime}}$ is the concrete compressive strength (4000 psi).
• ${b_w}$ is the beam width (15 inches).
• ${d}$ is the effective depth of the beam. To calculate 'd,' we subtract the concrete cover and half the stirrup diameter from the total depth. Assuming a concrete cover of 1.5 inches and No. 4 stirrups, ${d = 30 \text{ in} - 1.5 \text{ in} - 0.5 \text{ in} = 28 \text{ in}}$

Let's plug these values into the equation:

${ V_c = 2 \times 1.0 \times \sqrt{4000} \times 15 \times 28 = 53,257 \text{ lbs} = 53.26 \text{ kips} }$

So, the concrete shear capacity (${V_c}$) is approximately 53.26 kips. This represents the shear force that the concrete alone can resist. It's important to know this value so we can determine how much shear the stirrups need to handle.

Next, we need to determine the shear force that the stirrups must resist. This is simply the difference between the total nominal shear strength (${V_n}$) and the concrete shear capacity (${V_c}$):

${ V_s = V_n - V_c }$

Plugging in the values we calculated earlier:

${ V_s = 106.67 \text{ kips} - 53.26 \text{ kips} = 53.41 \text{ kips} }$

Thus, the required shear strength from stirrups (${V_s}$) is 53.41 kips. This is the shear force that the stirrups must be designed to resist, ensuring the beam's overall shear capacity.

Now, let's consider the torsional moment (${T_u}$) of 900 in-kips. We need to determine the torsional capacity provided by the concrete (${T_{cr}}$). This is the threshold below which torsional effects can be considered negligible. The formula for the cracking torsional moment (${T_{cr}}$) is:

${ T_{cr} = 4 \lambda \sqrt{f_c^{\prime}} \frac{A_{cp}^2}{p_{cp}} }$

Where:

• ${A_{cp}}$ is the area enclosed by the outside perimeter of the concrete cross-section.
• ${p_{cp}}$ is the outside perimeter of the concrete cross-section.

For our beam:

${ A_{cp} = b_w \times h = 15 \text{ in} \times 30 \text{ in} = 450 \text{ in}^2}$

${ p_{cp} = 2(b_w + h) = 2(15 \text{ in} + 30 \text{ in}) = 90 \text{ in}}$

Now, let's calculate ${T_{cr}}$:

${ T_{cr} = 4 \times 1.0 \times \sqrt{4000} \times \frac{450^2}{90} = 505,964 \text{ in-lbs} = 505.96 \text{ in-kips} }$

Now, we need to check if the factored torsional moment (${T_u}$) is greater than ${\phi \frac{T_{cr}}{4}}$, where ${\phi}$ is 0.75. If it is, we need to consider torsion in our design.

${ \phi \frac{T_{cr}}{4} = 0.75 \times \frac{505.96 \text{ in-kips}}{4} = 94.87 \text{ in-kips} }$

Since our factored torsional moment (${T_u}$) of 900 in-kips is much greater than 94.87 in-kips, we must design for torsion. This means we'll need to include additional stirrups to resist the torsional forces.

To handle the torsional effects, we need to calculate the required area of stirrups for torsion (${A_t/s}$). The ACI code provides the following formula:

${ \frac{A_t}{s} = \frac{T_u}{ \phi 0.85 A_o f_{yt}} }$

Where:

• ${T_u}$ is the factored torsional moment (900 in-kips).
• ${\phi}$ is the strength reduction factor for torsion (0.75).
• ${A_o}$ is the area enclosed by the torsional stirrups. We can approximate ${A_o}$ as 0.85 times the area enclosed by the centerline of the stirrups (${A_{oh}}$). Let's assume a 1.5-inch clear cover and No. 4 stirrups. Then:
  • The width inside the stirrups is ${b_{oh} = 15 \text{ in} - 2 \times (1.5 \text{ in} + 0.5 \text{ in}) = 11 \text{ in}}$
  • The depth inside the stirrups is ${h_{oh} = 30 \text{ in} - 2 \times (1.5 \text{ in} + 0.5 \text{ in}) = 26 \text{ in}}$
  • So, ${A_{oh} = b_{oh} \times h_{oh} = 11 \text{ in} \times 26 \text{ in} = 286 \text{ in}^2}$
  • And, ${A_o = 0.85 \times A_{oh} = 0.85 \times 286 \text{ in}^2 = 243.1 \text{ in}^2}$
• ${f_{yt}}$ is the yield strength of the stirrup steel (60,000 psi).

Now we can calculate ${A_t/s}$:

${ \frac{A_t}{s} = \frac{900 \text{ in-kips}}{0.75 \times 0.85 \times 243.1 \text{ in}^2 \times 60 \text{ ksi}} = 0.081 \frac{\text{in}^2}{\text{in}} }$

This value represents the required area of one leg of the stirrup per unit length to resist torsion.

Since stirrups resist both shear and torsion, we need to combine the reinforcement requirements. For shear, the stirrup spacing (${s}$) is determined by:

${ s = \frac{A_v f_{yt} d}{V_s} }$

Where:

• ${A_v}$ is the area of the stirrup resisting shear (for No. 4 stirrups, it's 2 legs \times 0.20 \text{ in}^2 = 0.40 \text{ in}^2)).
• ${f_{yt}}$ is the yield strength of the stirrup steel (60,000 psi).
• ${d}$ is the effective depth (28 inches).
• ${V_s}$ is the shear force resisted by stirrups (53.41 kips = 53,410 lbs).

So,

${ s = \frac{0.40 \text{ in}^2 \times 60,000 \text{ psi} \times 28 \text{ in}}{53,410 \text{ lbs}} = 12.56 \text{ in} }$

For torsion, we have ${A_t/s = 0.081 \text{ in}^2/\text{in}}$ and since stirrups have two legs, the area provided by the stirrups for torsion is 2${A_t}$ (because torsional stress acts on all faces of the section). Thus, the stirrup spacing required for torsion is:

${ s = \frac{2 A_t}{A_t/s} = \frac{2 \times 0.20 \text{ in}^2}{0.081 \text{ in}^2/\text{in}} = 4.94 \text{ in} }$

To satisfy both shear and torsion requirements, we need to use the smaller spacing. So, the required stirrup spacing based on torsion is approximately 4.94 inches.

ACI code also sets maximum spacing limits to ensure adequate shear and torsion reinforcement. The maximum spacing for shear is:

${ s_{max} = \frac{d}{2} = \frac{28 \text{ in}}{2} = 14 \text{ in} }$

And also:

${ s_{max} = 24 \text{ in} }$

The maximum spacing for torsion is:

${ s_{max} = \frac{p_h}{8}}$

Where ${p_h}$ is the perimeter of the centerline of the closed stirrups:

${ p_h = 2(b_{oh} + h_{oh}) = 2(11 \text{ in} + 26 \text{ in}) = 74 \text{ in}}$

So,

${ s_{max} = \frac{74 \text{ in}}{8} = 9.25 \text{ in} }$

Additionally, the ACI code specifies that the spacing should not exceed 12 inches when torsion reinforcement is required. This is a critical requirement to ensure the structure's torsional capacity and prevent cracking.

Considering all these limits, the maximum stirrup spacing is the smallest of these values, which is 4.94 inches (governed by torsion requirements). However, since we should use practical values, we can round down to 4.5 inches or even 4 inches to provide a bit of extra safety margin. This is a common practice in structural design to account for any unforeseen factors.

We also need to check for minimum stirrup requirements. The minimum area of shear reinforcement (${A_{v,min}}$) is given by:

${ A_{v,min} = 0.75 \sqrt{f_c^{\prime}} \frac{b_w s}{f_{yt}} \geq 50 \frac{b_w s}{f_{yt}} }$

Let's use a spacing of 4.5 inches and check:

${ A_{v,min} = 0.75 \sqrt{4000} \frac{15 \text{ in} \times 4.5 \text{ in}}{60,000 \text{ psi}} = 0.053 \text{ in}^2}$

And:

${ A_{v,min} = 50 \frac{15 \text{ in} \times 4.5 \text{ in}}{60,000 \text{ psi}} = 0.056 \text{ in}^2}$

So, the minimum ${A_{v,min}}$ is 0.056 ${\text{in}^2}$. For No. 4 stirrups at 4.5 inches spacing, the provided area is:

${ A_v = \frac{0.40 \text{ in}^2}{4.5 \text{ in}} = 0.089 \frac{\text{in}^2}{\text{in}} }$

Since 0.089 ${\text{in}^2}$ is greater than 0.056 ${\text{in}^2}$ and 0.053 ${\text{in}^2}$ we meet the minimum stirrup requirement. Always double-check these minimum requirements to ensure the structural integrity and longevity of the beam.

In addition to stirrups, we also need longitudinal reinforcement to resist torsion. The ACI code specifies that the additional longitudinal reinforcement (${A_l}$) should be calculated as:

${ A_l = \frac{A_t}{s} p_h \frac{f_{yt}}{f_y} }$

Where:

• ${A_t/s}$ is 0.081 ${\text{in}^2/\text{in}}$ (from our torsion calculation).
• ${p_h}$ is 74 inches.
• ${f_{yt}}$ is the yield strength of the stirrup steel (60,000 psi).
• ${f_y}$ is the yield strength of the longitudinal steel (60,000 psi). Assuming the same steel is used for longitudinal reinforcement.

Plugging in the values:

${ A_l = 0.081 \frac{\text{in}^2}{\text{in}} \times 74 \text{ in} \times \frac{60,000 \text{ psi}}{60,000 \text{ psi}} = 5.99 \text{ in}^2 }$

The longitudinal reinforcement should be distributed around the perimeter of the section. The ACI code also provides a minimum requirement for longitudinal torsional reinforcement, which should not be less than:

${ A_{l,min} = \frac{5 \sqrt{f_c^{\prime}} A_{cp}}{f_y} - \left(\frac{A_t}{s}\right) p_h \frac{f_{yt}}{f_y} }$

But ${\left(\frac{A_t}{s}\right) p_h \frac{f_{yt}}{f_y}}$ need not be taken greater than ${\frac{400 A_{cp}}{f_y}}$.

Let's calculate:

${ \frac{5 \sqrt{f_c^{\prime}} A_{cp}}{f_y} = \frac{5 \sqrt{4000} \times 450 \text{ in}^2}{60,000 \text{ psi}} = 2.37 \text{ in}^2 }$

And,

${ \left(\frac{A_t}{s}\right) p_h \frac{f_{yt}}{f_y} = 0.081 \frac{\text{in}^2}{\text{in}} \times 74 \text{ in} \times \frac{60,000 \text{ psi}}{60,000 \text{ psi}} = 5.99 \text{ in}^2 }$

Also,

${ \frac{400 A_{cp}}{f_y} = \frac{400 \times 450 \text{ in}^2}{60,000 \text{ psi}} = 3 \text{ in}^2 }$

So, we use 3 ${\text{in}^2}$ as the upper limit for the second term in the minimum longitudinal reinforcement calculation:

${ A_{l,min} = 2.37 \text{ in}^2 - 3 \text{ in}^2}$

Since this results in a negative value, we take ${A_{l,min}}$ as zero (as we can't have a negative area of steel). Therefore, the required longitudinal reinforcement is governed by our initial calculation of 5.99 ${\text{in}^2}$.

We need to distribute this reinforcement around the perimeter of the beam. A common approach is to place bars in each corner of the stirrups and additional bars along the sides to meet the required area. This ensures that the torsional forces are adequately resisted along the entire section. For example, you might use four No. 9 bars (1 ${\text{in}^2}$ each) in the corners and two No. 6 bars (0.44 ${\text{in}^2}$ each) on each side, providing a total area of approximately 5.76 ${\text{in}^2}$ which is close to our calculated requirement. This distribution helps in effectively resisting torsional stresses and preventing cracking.

Alright, guys, we've crunched the numbers, considered all the code requirements, and have a final design! To recap, for a rectangular beam 15 inches wide and 30 inches deep, subjected to a factored shear of 80 kips and a factored torsional moment of 900 in-kips, with concrete strength (${f_c^{\prime}}$) of 4000 psi and steel yield strength (${f_{yz}}$) of 60,000 psi, we need:

• No. 4 closed stirrups spaced at 4.5 inches on center. This spacing satisfies both shear and torsion requirements while staying within ACI code limits.
• Longitudinal reinforcement of approximately 5.99 square inches, distributed around the perimeter of the beam. A practical arrangement could be four No. 9 bars in the corners and additional bars along the sides to meet this requirement.

Remember, these calculations are based on specific code provisions and assumptions. Always consult the latest ACI code and engineering best practices for your specific project. Structural design is a critical aspect of any construction project, and it's essential to get it right. The safety and durability of the structure depend on accurate calculations and adherence to industry standards.

It's also important to consider practical aspects, such as constructability and bar placement. Sometimes, a slightly adjusted design might be easier to build in the field, without compromising structural integrity. This collaborative approach between design engineers and construction teams often leads to the most efficient and cost-effective solutions.

I hope this detailed walkthrough has been helpful in understanding how to calculate stirrup spacing for combined shear and torsion. Keep practicing, and you'll become a pro in no time! Happy designing!