框架混凝土梁粘贴钢板加固如何进行设计计算?

  公司新闻     |      2025-03-18 05:14

框架混凝土梁粘贴钢板加固如何进行设计计算?下面贵州新东源建筑加固工程有限公司小编给大家试算框架混凝土梁粘贴钢板加固如何进行设计计算。

一、梁设计条件:

截面尺寸 b =250 mm , h =600 mm ,,混凝土强度等级C30 ,底筋主筋3Φ18(AS=763 mm2)(底筋主筋4Φ18(AS=1017 mm2)。

IMG_20240101_143604

1、梁承载力复核(双筋梁正截面复核)

1)L1荷载作用产生的弯矩设计值:

b= 250; h= 600; h0= 565.0mm,非抗震设计                                                                            

C35; Fc= 16.7; α1=1.000; Fy= 300.MPa; As=   763.00; A"s=   401.00 mm^2                                              

ρ= 0.0054, ρmin= 0.0024                                                                                             

x=(fy×As-f"y×A"s)/(α1×fc×b)= 300.×( 763.0- 401.0)/(1×16.7× 250)= 26.0mm                                      

算出的x=   26.0mm, 取x=2a"= 70.00mm                                                                                  

按照《混凝土结构设计规范》式(6.2.14), Mu=FyAs(h0-a")                                                                  

双筋梁能承担弯矩 Mu=    121.3 kN.m              

IMG_20231230_100248                                                                                                              

2) L2荷载作用产生的弯矩设计值:

b= 250; h= 600; h0= 565.0mm,非抗震设计                                                                             

C30; Fc= 14.3; α1=1.000; Fy= 300.MPa; As= 1017.00; A"s=   401.00 mm^2                                              

ρ= 0.0072, ρmin= 0.0021                                                                                              

x=(fy×As-f"y×A"s)/(α1×fc×b)= 300.×(1017.0- 401.0)/(1×14.3× 250)= 51.7mm                                      

算出的x=   51.7mm, 取x=2a"= 70.00mm                                                                                   

按照《混凝土结构设计规范》式(6.2.14), Mu=FyAs(h0-a")                                                                  

双筋梁能承担弯矩 Mu=    161.7 kN-m    

二、加固设计

增加弯矩计算:


1)受拉面粘贴钢板计算,非抗震设计   

b= 250mm, h= 600mm, b"f= 400mm, h"f=   80mm, a= 40mm, a"= 40mm                                           

x=fy(As-A"s)/(α1Fcb"f)= 36.02mm; 中和轴在翼缘内                                                           

Mu0=α1×Fc×b"f×x(h0-0.5x)=1.00×14.3× 400× 36.02×( 560-0.5× 36.02)= 111.66kN-m                     

A= 2860.0, B= -3431999.8, C=    162540400., x=    49.4                                                     

ξ≤0.85ξb=0.4896, 满足要求。                                                                              

不考虑二次受力影响, 取εsp0=0.0                                                                             

ψsp=(0.8×0.0033×h/x-0.0033-εsp0)/(Fpy/Esp)= 27.565                                                      

ψsp>1, 取ψsp=1                                                                                            

x≤h"f, 属于第一类T形截面                                                                                    

Asp=(α1×fc0×b"f×x-fy0×As0+f"y0×A"s0+f"sp×A"sp)/(ψsp×fsp)                                           

原梁能承担Mu= 111.7kN-m,提高 38.2%,受拉钢板Asp= 355.9mm^2                                                 

可选q235钢板200×2=400 mm2

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