题目
9-5 阶梯轴AB如图所示,AC段直径 _(1)=40mm, CB段直径 _(2)=70mm, 外力偶矩 _(B)=1500Ncdot m ,MA=600N·m,-|||-_(c)=900Ncdot m-|||-,G=80GPa,-|||-[ t] =60MPa,-|||-MA Mc MB-|||-[ varphi '] =2(0)|m 试校核该轴的强度和刚度。-|||-MA 1. Mc 2. MB-|||-A 1 C 2 B-|||-↑T 1500N·m-|||-600N·m +-|||-x-|||-A c B
题目解答
答案
解析
步骤 1：计算扭矩
首先，根据题目中给出的外力偶矩，计算出各段轴的扭矩。对于AC段，扭矩为 ${T}_{1} = {M}_{A} = 600N\cdot m$；对于CB段，扭矩为 ${T}_{2} = {M}_{B} - {M}_{C} = 1500N\cdot m - 900N\cdot m = 600N\cdot m$。根据计算结果，画出扭矩图。
步骤 2：校核轴的强度
根据扭转强度条件，计算各段轴的最大切应力。对于AC段，直径 ${d}_{1} = 40mm$，最大切应力 ${t}_{max1} = \dfrac {16{T}_{1}}{\pi {d}_{1}^{3}} = \dfrac {16 \times 600N\cdot m}{\pi \times {0.04}^{3}m^{3}} = 30.2MPa$；对于CB段，直径 ${d}_{2} = 70mm$，最大切应力 ${t}_{max2} = \dfrac {16{T}_{2}}{\pi {d}_{2}^{3}} = \dfrac {16 \times 600N\cdot m}{\pi \times {0.07}^{3}m^{3}} = 13.5MPa$。比较计算结果与许用切应力 $[t] = 60MPa$，判断轴的强度是否满足要求。
步骤 3：校核轴的刚度
根据扭转刚度条件，计算各段轴的单位长度扭转角。对于AC段，极惯性矩 ${I}_{p1} = \dfrac {\pi {d}_{1}^{4}}{32} = \dfrac {\pi \times {0.04}^{4}m^{4}}{32} = 251 \times {10}^{-9}m^{4}$，单位长度扭转角 ${\phi}_{1} = \dfrac {{T}_{1}}{G{I}_{p1}} = \dfrac {600N\cdot m}{80 \times {10}^{9}Pa \times 251 \times {10}^{-9}m^{4}} = 0.00301rad/m$；对于CB段，极惯性矩 ${I}_{p2} = \dfrac {\pi {d}_{2}^{4}}{32} = \dfrac {\pi \times {0.07}^{4}m^{4}}{32} = 2.36 \times {10}^{-6}m^{4}$，单位长度扭转角 ${\phi}_{2} = \dfrac {{T}_{2}}{G{I}_{p2}} = \dfrac {600N\cdot m}{80 \times {10}^{9}Pa \times 2.36 \times {10}^{-6}m^{4}} = 0.00313rad/m$。比较计算结果与许用单位长度扭转角 $[\phi] = 2(0)/m$，判断轴的刚度是否满足要求。
首先，根据题目中给出的外力偶矩，计算出各段轴的扭矩。对于AC段，扭矩为 ${T}_{1} = {M}_{A} = 600N\cdot m$；对于CB段，扭矩为 ${T}_{2} = {M}_{B} - {M}_{C} = 1500N\cdot m - 900N\cdot m = 600N\cdot m$。根据计算结果，画出扭矩图。
步骤 2：校核轴的强度
根据扭转强度条件，计算各段轴的最大切应力。对于AC段，直径 ${d}_{1} = 40mm$，最大切应力 ${t}_{max1} = \dfrac {16{T}_{1}}{\pi {d}_{1}^{3}} = \dfrac {16 \times 600N\cdot m}{\pi \times {0.04}^{3}m^{3}} = 30.2MPa$；对于CB段，直径 ${d}_{2} = 70mm$，最大切应力 ${t}_{max2} = \dfrac {16{T}_{2}}{\pi {d}_{2}^{3}} = \dfrac {16 \times 600N\cdot m}{\pi \times {0.07}^{3}m^{3}} = 13.5MPa$。比较计算结果与许用切应力 $[t] = 60MPa$，判断轴的强度是否满足要求。
步骤 3：校核轴的刚度
根据扭转刚度条件，计算各段轴的单位长度扭转角。对于AC段，极惯性矩 ${I}_{p1} = \dfrac {\pi {d}_{1}^{4}}{32} = \dfrac {\pi \times {0.04}^{4}m^{4}}{32} = 251 \times {10}^{-9}m^{4}$，单位长度扭转角 ${\phi}_{1} = \dfrac {{T}_{1}}{G{I}_{p1}} = \dfrac {600N\cdot m}{80 \times {10}^{9}Pa \times 251 \times {10}^{-9}m^{4}} = 0.00301rad/m$；对于CB段，极惯性矩 ${I}_{p2} = \dfrac {\pi {d}_{2}^{4}}{32} = \dfrac {\pi \times {0.07}^{4}m^{4}}{32} = 2.36 \times {10}^{-6}m^{4}$，单位长度扭转角 ${\phi}_{2} = \dfrac {{T}_{2}}{G{I}_{p2}} = \dfrac {600N\cdot m}{80 \times {10}^{9}Pa \times 2.36 \times {10}^{-6}m^{4}} = 0.00313rad/m$。比较计算结果与许用单位长度扭转角 $[\phi] = 2(0)/m$，判断轴的刚度是否满足要求。