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物料提升机计算书

上传者:菩提 |  格式:doc  |  页数:5 |  大小:106KB

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; M 1=2.1 2/12 ×[(2 ×4.8+0.9) ×(14.659+14.659-2 × 5 77.76/11.52)+(14.659-14.659) ×4.8]=61.037kN ·m; M 2=(4.8-0.9) 2/48 ×(2×2.4+0.6) ×(14.659+14.659-2 × 77.76/11.52)=27.066kN ·m; 基础配筋计算式中 a 1---- 砼强度等级不超过 C50 ,取 a 1=1; 1-1 截面: α s =|M|/(a 1f c bh 0 2 )=61.04 × 10 6 /(1.00 × 9.60 × 4.80 × 10 3× 165.00 2 )=0.049 ; ξ=1-(1- α s) 1/2 =1-(1-2 × 0.049) 0.5 =0.050 ; γ s =1- ξ/2=1-0.050/2=0.975 ; A s=|M|/( γ sf yh 0 )=61.04 × 10 6 /(0.975 × 210.00 × 165.00)=806.59mm 2。 2-2 截面:α s =|M|/(a 1f c bh 0 2 )=27.07 × 10 6 /(1.00 × 9.60 × 2.40 × 10 3× 165.00 2 )=0.043 ; ξ=1-(1- α s) 1/2 =1-(1-2 × 0.043) 0.5 =0.044 ; γ s =1- ξ/2=1-0.044/2=0.978 ; A s=|M|/( γ sf yh 0 )=27.07 × 10 6 /(0.978 × 210.00 × 165.00)=798.75mm 2。截面 1-1 配筋:A s1=1067.566 mm 2≥806.59 mm 2 截面 2-2 配筋:A s2=906.858 mm 2≥798.749 mm 2 承台配筋满足要求!

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