简支梁承受最大弯矩怎么计算?
设:梁长L;均布荷载Q;跨中最大弯矩M.
取跨中为平衡点,此时有:
支座反力:大小为QL/2,方向向上(为正),作用点距离L/2.
半跨均布荷载:大小QL/2,方向向下(为负),作用点距离L/4.取矩则有:
M=QL/2*L/2(支座反力作用)-QL/2*L/4(半跨均布荷载作用)=1/8*QL2.
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