一矩形截面钢筋混凝土梁，材料采用C30混凝土，钢筋采用三级钢，梁截面尺寸b×h=250×550m㎡

Question

一矩形截面钢筋混凝土梁，材料采用C30混凝土，钢筋采用三级钢，梁截面尺寸b×h=250×550m㎡求所需配置的箍筋。已知：as=35mm,受集中剪力值为180KN，剪跨为1.5m

Answer 1

可给你一个参考计算过程：

梁正截面受弯承载力计算书

1 已知条件

    æ¢æˆªé¢å®½åº¦b=250mm,高度h=600mm,受压钢筋合力点至截面近边缘距离a^'_s=35mm,受拉钢筋合力点到截面近边缘距离a_s=35mm,计算跨度l₀=6300mm,混凝土强度等级C20,纵向受拉钢筋强度设计值f_y=300MPa,纵向受压钢筋强度设计值f^'_y=300MPa,非抗震设计,设计截面位于框架梁梁中,截面设计弯矩M=142.88kN·m,截面下部受拉。

2 配筋计算

    æž„件截面特性计算

                       A=150000mm², I_x=4499999744.0mm⁴

    æŸ¥æ··å‡åœŸè§„范表4.1.4可知

                       f_c=9.6MPa        f_t=1.10MPa

    ç”±æ··å‡åœŸè§„范6.2.6条可知

                       Î±₁=1.0             Î²₁=0.8

    ç”±æ··å‡åœŸè§„范公式(6.2.1-5)可知混凝土极限压应变

                       Îµ_cu=0.0033

    ç”±æ··å‡åœŸè§„范表4.2.5可得钢筋弹性模量

                       E_s=200000MPa

    ç›¸å¯¹ç•Œé™å—压区高度

                       Î¾_b=0.550

    æˆªé¢æœ‰æ•ˆé«˜åº¦

                       h₀=h-a^'_s=600-35=565mm

    å—拉钢筋最小配筋率

                       Ï_smin=0.0020

    å—拉钢筋最小配筋面积

                       A_smin=ρ_sminbh

                          =0.0020×250×600

                          =300mm²

    æ··å‡åœŸèƒ½æ‰¿å—的最大弯矩

                       M_cmax=α₁f_cξ_bh₀b(h₀-0.5ξ_bh₀)

                          =1.0×9.6×0.550×565×250×(565-0.5×0.550×565)

                          =304043584N·mm >M

    ç”±æ··å‡åœŸè§„范公式(6.2.10-1)可得

                       Î±_s=M/α₁/f_c/b/h²₀

                         =142880000/1.0/9.6/250/565²

                         =0.19

    æˆªé¢ç›¸å¯¹å—压区高度

                       Î¾=1-(1-2α_s)^0.5=1-(1-2×0.19)^0.5=0.209

    ç”±æ··å‡åœŸè§„范公式(6.2.10-2)可得受拉钢筋面积

                       A_s=(α₁f_cbξh₀)/f_y

                         =(1.0×9.6×250×0.21×565)/300

                         =941.47mm²

    A_s>A_smin,取受拉钢筋面积

                       A_s=941.47mm²

梁斜截面受剪承载力计算书

 

1 已知条件

    æ¢æˆªé¢å®½åº¦b=250mm,高度h=600mm,纵向钢筋合力点至截面近边缘距离a_s=35mm,计算跨度l₀=6300mm,箍筋间距s=100mm,混凝土强度等级C20,箍筋设计强度f_yv=270MPa,非抗震设计,竖向剪力设计值V=90.72kN,求所需钢筋面积。

2 配筋计算

    æŸ¥æ··å‡åœŸè§„范表4.1.4可知

                       f_c=9.6MPa        f_t=1.10MPa

    ç”±æ··å‡åœŸè§„范6.3.1条可得混凝土强度影响系数

                       Î²_c=1.0

    æˆªé¢é¢ç§¯

                       A=bh

                        =250×600

                        =150000mm²

    æˆªé¢æœ‰æ•ˆé«˜åº¦

                       h₀=h-a_s=600-35=565mm

    æˆªé¢è…¹æ¿é«˜åº¦

                       h_w=h₀=565mm

    ç”±æ··å‡åœŸè§„范6.3.1条可知截面允许的最大剪应力

                       Ï„_max=0.25β_cf_c=0.25×1.0×9.6=2.39MPa

    å‰ªåŠ›äº§ç”Ÿçš„剪应力

                       Ï„_V=V/b/h₀

                         =90720/250/565

                         =0.64MPa

    å‰ªåº”力τ_V<τ_max,截面尺寸满足条件。

    ç”±æ··å‡åœŸè§„范可知混凝土的抗剪承载力

                       V_c=0.7f_tbh₀

                         =0.7×1.10×250×565

                         =108889.16N

    ç”±æ··å‡åœŸè§„范6.3.13可知

                       V/b/h₀=0.64MPa

                       V_c/b/h₀=0.77MPa

    åˆ™å¯ä¸è¿›è¡Œæ‰¿è½½åŠ›è®¡ç®—，取

                       A_sv=0

    å–箍筋最小配筋率

                       Ï_svmin=0.0006

    åŒä¸€æˆªé¢æœ€å°ç®ç­‹é¢ç§¯

                       A_svmin=ρ_svminbs

                           =0.0006×250×100

                           =16.17mm² >A_sv

    å–箍筋面积

                       A_sv=16.17mm²