精选优质文档-----倾情为你奉上1. 平衡梁校核说明：1）2）按吊装载荷10t校核3）跨距为2m4）梁下索具为垂直承力5）梁上索具使用偏角不小于60°（该计算书按60°校核）6）设计安全系数为1.62. 平衡梁示意图3. 与校核G=36.75561kg按吊装载荷10t校核平衡梁在动态平衡状态下，受力如上图示简化，见下图：由作图法知F1 = F4 = 25.882kN平衡梁处于动态平衡中，竖直方向力相等所以：F1×sin15°+ F4×sin15°+ G = F2×sin74°+ F3×sin74°又： F2 = F3所以：F2 = F3 = 7.156kN所以平衡梁受力如图对平衡梁进行受力分析平衡梁轴向力方程：F1横 + F2横 = F3横 + F4横F1横 = F4横 = F1×cos15°= 25.000kNF2横 = F3横 = F2×cos74°= 1.972kN平衡梁纵向力方程：F1竖 = F4竖 = F1×sin15°= 6.699kNF2竖 = F3竖 = F2×sin74°= 6.879kN1）做剪力图、弯矩图（1）剪力图AB、BC、CD、DE各段无载荷作用，FQ为常数，FQ图为斜直线。
从A端开始，求出控则截面剪力如下：FQA = -F1竖 = -6.699kNFQB = F2竖 + FQA = 6.879 - 6.699 = 0.18 kNFQC = - G +FQB = - 0.360 + 0.18= - 0.18 kNFQD = F3竖+FQC = 6.879 -0.18= 6.699 kN（2）弯矩图选A、B、C、D、E为控制截面，求出其弯矩值如下：MA = 0MB = F1竖×0.5 = 6.699×0.5 =3.350kN·mMC = F1竖×1 – F2竖×0.5 = 6.699×1 - 6.879×0.5 = 3.260kN·mMD = F1竖×1.5–F2竖×1+G×0.5=6.699×1.5-6.879×1+0.360×0.5=3.350kN·mME = 02）刚度校核由《起重机设计手册》P521式4-1-30刚度条件为：λ≤［λ］λ=μL / r 式中：L=2000mm r = (I / A)1/2 I = π(D4 – d4 ) / 64对于Φ114×6钢管，I =π(1144 – 1024 ) / 64 = 2.977×106 mm4A =π(572 – 512 ) = 2035.752 mm2故 r = (I / A)1/2 =38.243mm查《起重机设计手册》P545表4-1-40取 μ = 1查《起重机设计手册》P522表4-1-16取［λ］= 150所以 λ=1×2000/38.243 = 52.297 ≤［λ］ 所以支撑梁刚度满足要求。
3）强度校核由《起重机设计手册》P521式4-1-30刚度条件为：σ= F / A ≤［σ］式中：F = G/2 = 4.9×104×tan30o = 2.829×104N A = 2035.752 mm2故 σ= 13.897 MPa钢管材料为20钢，σ= 13.897 MPa ≤［σ］= 245 Mpa 所以强度满足要求4）稳定性校核由《起重机设计手册》P523式4-1-36稳定性条件为：σ= F / (ψA) ≤［σ］式中：F = G/2 = 4.9×104×tan30o = 2.829×104N A = 2035.752 mm2由《起重机设计手册》P523表4-1-17取ψ = 0.847所以 σ= 16.407 MPa ≤［σ］所以稳定性满足要求4. 吊耳校核吊耳材质为Q235，σs=235 Mpa安全系数为1.6[σ] =σs/1.6 = 146.875 Mpa[τ] = 80.781 Mpa (取0.55[σ])1）吊耳截面拉伸应力计算吊耳拉伸载荷F2 = 7.156kN吊耳截面拉应力σ=Fv /A式中：Fv = F2 = 7.156kN A = (50-30) ×5 = 100 mm2故σ=F2 /A=7.156×103/[(50-30)×5]=71.560Mpa ≤[σ] = 125 Mpa强度满足。
2）吊耳孔挤压强度计算σr =式中：T = F2 = 7.156kNd 为吊耳孔直径，d = 30mmt1 为吊耳板厚度，t1 = 5mmt2 为加强板厚度，t1 = 0mm故σr =110 Mpa ≤[σ] = 125 Mpa强度满足3）吊耳截面剪切应力计算τ=式中：Fv = F2 = 7.156kN A = (50-30) ×5 = 100 mm2τ= 71.560Mpa ≤ [τ] = 80.781 Mpa强度满足 通过以上分析计算证实该平衡梁能够满足要求专心---专注---专业。