Static condition

\begin{aligned} &\sum{M}=0\\ &M_A+M_D+20+\frac{1}{2}\cdot 15*3-5\cdot 2=0\\ &M_A+M_D+32.5=0\\ \end{aligned}

Geometric condition \Delta\varphi_{c}=0

Torque moments in characteristic intervals

\begin{aligned} &x\in (0;2)\\ &M_{AB}=M_A-5x\\ &x\in (0;3)\\ &M_{BC}=M_A-10+20=M_A+10\\ \\ &x\in (0;3)\\ &\frac{15}{3}=\frac{m}{x_3}\\ &m=5x_3\\ &M_{CD}=M_A+10+\frac{1}{2}\cdot m\cdot x_3=M_A+10+2.5x_3^2\\ \end{aligned}

Geometric condition

\begin{aligned} &\int\limits_{0}^{2}\frac{M_A-5x}{G*I_S}dx + \int\limits_{0}^{3}\frac{M_A+10}{G*I_S}dx + \int\limits_{0}^{3}\frac{M_A+10+2.5x^2}{G*I_S}dx=0\ |\cdot GI_S\\ &2M_A-\frac{5}{2}*2^2+3M_A+30+3M_A+30+2.5*\frac{3^3}{3}=0\\ & M_A=-9.0625 \ kNm\\ \\ &M_{AB}=-9.0625-5x \\ & M_{A(0)}=-9.0625 \ kNm \\ & M_{B(2)}=-19.0625\ kNm\\ \\ &M_{BC}=-9.0625+10=0.9375 \ kNm\\ \\ &M_{CD}=-9.0625+10+2.5*x^2 \\ & M_{C(0)}= 0.9375\ kNm \\ & M_{D(3)}=21.09375 \ kNm\\ \end{aligned}

Torque moments diagram

Designing the rod diameter

(1) Service condition

\begin{aligned} &M_{max}=21.09375\ kNm\\ &\Theta_{max}\le\Theta_{dop} \\ &\frac{M_S}{G*I_S}\le\Theta_{dop} \\ & I_S=\frac{\pi*d^4}{32}\\ &\frac{21.09275*10^3}{80*10^9*\frac{\pi*d^4}{32}}\le 0.3*\frac{\pi}{180}\\ &d\ge 0.15049\ m\\ \end{aligned}

(2) Safety condition

\begin{aligned} &\tau_{max}\le k_s \\ &\frac{M_S}{W_S}\le k_s \\ & W_S=\frac{\pi * d^3}{16}\\ &\frac{21.09375*10^3}{\frac{\pi*d^3}{16}}\le70*10^6 \\ &d\ge 0.0392\ m \\ \end{aligned}

I design the rod diameter: d=0.16 m

Unit angle of twist

\begin{aligned} &\Theta=\frac{M_S}{G*I_S} \\ & G*I_S=80*10^9*\frac{\pi*(0.16)^4}{32}=5147.1854*10^3\ Nm^2\\ &\Theta_A=-0.1 \ [^\circ/m] \\ &\Theta_{B^L}=-0.212 \ [^\circ/m]\\ & \Theta_{B^P}=0.01 \ [^\circ/m]\\ &\Theta_C=0.01 \ [^\circ/m] \\ & \Theta_D=0.2348 \ [^\circ/m]\\ \end{aligned}

Angle of twist

\begin{aligned} &\Delta\varphi_{AB}=\int\limits_{0}^{2}\frac{M_A-5x}{G*I_S}dx=\frac{1}{G*I_S}*(-9.0625*2-\frac{5}{2}*4)=-0.31307 \ [^\circ]\\ &\Delta\varphi_{BC}=\int\limits_{0}^{3}\frac{M_A+10}{G*I_S}dx=\frac{1}{G*I_S}*(-9.0625*3+30)=0.031307 \ [^\circ]\\ &\Delta\varphi_{CD}=\int\limits_{0}^{3}\frac{M_A+10+2.5x^2}{G*I_S}dx=\frac{3M_A+30+2.5\frac{3^3}{3}}{G*I_S}=0.28176 \ [^\circ]\\ \\ &\varphi_A=0 \\ & \varphi_B=\varphi_A+\Delta\varphi_{AB}=-0.31307\ [^\circ] \\ & \varphi_C=\varphi_B+\Delta\varphi_{BC}=-0.28176 \ [^\circ]\\ &\varphi_D=\varphi_C+\Delta\varphi_{CD}\simeq 0\\ \end{aligned}

Charts

Torsion

Strength of materials - Torsion - Example 8

Example 8