 SkyCiv文档

1. 讲解
2. 钢筋混凝土教程
3. 如何设计钢筋混凝土梁?

# 如何设计钢筋混凝土梁?

If you are new to beam design, we would recommend reading some introductory SkyCiv articles:

These tutorials will help you gain a better understanding of the general process of designing beams.

## SkyCiv Beam软件

The first stop is creating the beam model in the SkyCiv Beam Software. We indicate the steps required: (In parenthesis, we show the example data):

• On the dashboard page, 选择光束模块.
• 创建定义其长度的梁 (66 英尺).
• 转到支撑并定义铰链或简单的杆 (hinge at the beginning and the end; 第三点杆).
• 转到部分并创建一个矩形部分 (矩形截面; 宽度=18 英寸; 高度=24 英寸).
• 然后选择分布式负载按钮并分配一个, 二, or more as you need for (叠加恒载 = 0.25 基普/英尺; 活荷载 = 0.40 基普/英尺)
• The next step is create some load combinations ($${L_d = 1.2\times D + 1.6\times L}$$)
• 最后, solve the beam! After solving the beam, we can check the results, like the bending diagram, to get their maximum values along the element length. The following images show the final output. The SkyCiv Beam Software gives us a table with the maximum values for forces, 压力, and displacement: Now is the time to select the design tab and select and define the input as reinforcement layout, analysis sections, some coefficients, 荷载组合, 等等. Look at figures 4 和 5 for more description.  Once all the data is ready, we can click the “检查一下” 纽扣. This action will give us then the results and the capacity ratios for strength and serviceability. ## SkyCiv结构3D

Now is the time to use Structural 3D! We recommend just returning to the beam software and clicking on the “在S3D中打开” 纽扣. This will help us prepare the model and its inputs in S3D.

Once we clicked the change button, the model was automatically created. Remember to save it! (If you need to familiarize yourself with this module, look at this tutorial link!) Now go directly to the “解决” icon choosing theLinear analysis” 选项. Feel free to check and compare results; 我们将使用 “设计” 选项. It is time to define all the characteristics required to evaluate the beam on the different tabs.  SkyCiv can check for a particular defined RC layout or calculate a section reinforcement optimization. We’d like to suggest you run this latter option.   ## ACI-318 Approximate Equations

When designing a continuous beam, ACI-318 permits using moment coefficients for bending calculations. (For more examples, feel free to visit these SkyCiv’s articles about 楼板设计)

Moments at critical sections are calculated with: $$M_u = coefficient \times w_u \times l_n^2$$. Where the coefficient can be obtained from the following:

• Exterior span:
• Negative exterior: $$\压裂{1}{16}$$
• Positive midspan: $$\压裂{1}{14}$$
• Negative interior:$$\压裂{1}{10}$$
• Interior span:
• 消极的: $$\压裂{1}{11}$$
• Positive midspan: $$\压裂{1}{16}$$

We’ll select two cases: the absolute maximum value for positive and negative bending moments.

$$wu=1.2\times D + 1.6\times L = 1.2 \次 0.25 + 1.6 \次 0.4 = 0.94 \压裂{基普}{英尺}$$

$$M_{ü,neg} = {\压裂{1}{10}}{\次 0.94 {\压裂{基普}{英尺}}}{\次 {(22 英尺)}^ 2} = 45.50 {基普}{英尺}$$

$$M_{ü,pos} = {\压裂{1}{14}}{\次 0.94 {\压裂{基普}{英尺}}}{\次 {(22 英尺)}^ 2} = 32.50 {基普}{英尺}$$

Flexure resistance calculation for negative moment, $${M_{ü,neg} = 45.50 {基普}{英尺}}$$

• 假设张力控制部分. $${\phi_f = 0.9}$$
• Beam width, $${b=18 in}$$
• 钢筋面积, $${A_s = \frac{亩}{\phi_f\times 0.9d\times fy}= frac{45.50 kip-ft \times 12 in -ft }{0.9\次 0.9(17 在 )\次 60 KSI}=0.66 {在}^ 2}$$
• $${\o{分} = 0.003162}$$. 钢材最小配筋面积, $${一个_{s,分}=\rho_{分}\times b\times d = 0.003162 \次 18 in \times 17 in =0.968 {在}^ 2}$$. 现在, 检查该部分是否表现为张力控制.
• $${a = frac{A_s\times f_y}{0.85\times f’c\times b} = frac{0.968 {在}^2\times 60 KSI}{0.85\次 4 ksi\times 18 在 }= 0.95 在}$$
• $${c = \frac{一个}{\测试版_1}= 分数{0.95 在}{0.85} = 1.12 在 }$$
• $${\变体_t = (\压裂{0.003}{C})\次 {(d – C)} = (\压裂{0.003}{1.12 在})\次 {(17在 – 1.12 在)} = 0.0425 > 0.005 }$$ 好的!, 这是一个张力控制部分!.

Flexure resistance calculation for positive moment, $${M_{ü,pos} = 32.50 {基普}{英尺}}$$

• 假设张力控制部分. $${\phi_f = 0.9}$$
• Beam width, $${b=18 in}$$
• 钢筋面积, $${A_s = \frac{亩}{\phi_f\times 0.9d\times fy}= frac{32.50 kip-ft \times 12 in -ft }{0.9\次 0.9(17 在 )\次 60 KSI}=0.472 {在}^ 2}$$
• $${\o{分} = 0.003162}$$. 钢材最小配筋面积, $${一个_{s,分}=\rho_{分}\times b\times d = 0.003162 \次 18 in \times 17 in =0.968 {在}^ 2}$$. 现在, 检查该部分是否表现为张力控制.
• $${a = frac{A_s\times f_y}{0.85\times f’c\times b} = frac{0.968 {在}^2\times 60 KSI}{0.85\次 4 ksi\times 18 在 }= 0.95 在}$$
• $${c = \frac{一个}{\测试版_1}= 分数{0.95 在}{0.85} = 1.12 在 }$$
• $${\变体_t = (\压裂{0.003}{C})\次 {(d – C)} = (\压裂{0.003}{1.12 在})\次 {(17在 – 1.12 在)} = 0.0425 > 0.005 }$$ 好的!, 这是一个张力控制部分!.