Solution
It is worth familiarizing oneself with the theoretical introduction regarding circular arches. This introduction is based on a partial, more detailed discussion of this example.
Equations of static equilibrium \begin{aligned} &\sum{x}=0 & H_A-4,698=0 && H_A=4,698 \ kN\\ &\sum{M_A}=0 & 3\cdot 4\cdot 2+1,71\cdot 6-4,698\cdot 2\sqrt3-V_B\cdot 8=0 && V_B=2,248 \ kN\\ &\sum{y}=0 & V_A-3\cdot 4-1,71+2,248=0 && V_A=11,462 \ kN\\ \end{aligned}
Projecting the reaction VB onto the normal and shear components to the mental section of the arch.
We express the variables x and y in terms of the angle alpha.
\begin{aligned} &\frac{x}{4}=cos\alpha\\ &x=4cos\alpha\\ &\frac{y}{4}=sin\alpha\\ &y=4sin\alpha\\ \end{aligned} Internal force functions in the first interval.
\begin{aligned} &N(\alpha)=-2,248 cos\alpha\\ &T(\alpha)=-2,248 sin\alpha\\ &M(\alpha)=2,248 (4-x)=8,992-8,992 cos\alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[{ }^0\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 0 & 0.000 & 1.000 & 2.248 & 0.000 & 0.000 \\ \hline 15 & 0.259 & 0.966 & 2.171 & -0.582 & 0.306 \\ \hline 30 & 0.500 & 0.866 & 1.947 & -1.124 & 1.205 \\ \hline 45 & 0.707 & 0.707 & 1.590 & -1.590 & 2.634 \\ \hline 60 & 0.866 & 0.500 & 1.124 & -1.947 & 4.496 \\ \hline \end{array}
Internal force functions in the second interval.
\begin{aligned} &N(\alpha)=-2,248 cos\alpha+1,71 cos\alpha-4,698 sin\alpha=-0,538 cos\alpha-4,698 sin\alpha\\ &T(\alpha)=-2,248\cdot sin\alpha+1,71\cdot sin\alpha+4,698 cos\alpha=-0,583 sin\alpha+4,698 cos\alpha\\ &M(\alpha)=2,248\cdot (4-x)-1,71\cdot (2-x)-4,698(y-2\sqrt3)=\\ &=8,992-8,992\cdot cos\alpha-3,42+6,84\cdot cos\alpha-18,792 sin\alpha+16,274=\\ &=21,848-2,152 cos\alpha-18,792 sin\alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[^{\circ}\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 60 & 0.866 & 0.500 & -4.338 & 1.883 & 4.496 \\ \hline 75 & 0.966 & 0.259 & -4.677 & 0.696 & 3.139 \\ \hline 90 & 1.000 & 0.000 & -4.698 & -0.538 & 3.056 \\ \hline \end{array}
Internal force functions in the third interval.
\begin{aligned} &N(\alpha)=-11,462\cdot cos\alpha -4,698\cdot sin\alpha+3\cdot (4-x) cos\alpha=\\ &=-11.462\cdot cos\alpha-4,698\cdot sin\alpha+12\cdot cos\alpha-12cos^2 \alpha=\\ &=0,538\cdot cos\alpha-4,698\cdot sin\alpha-12\cdot cos^2 \alpha\\ \\ &T(\alpha)=11,462\cdot sin\alpha-4,698\cdot cos\alpha-3\cdot (4-x)\cdot sin\alpha=\\ &=11,462\cdot sin\alpha-4,698\cdot cos\alpha-12\cdot sin\alpha+12\cdot sin\alpha cos\alpha=\\ &=-0,538\cdot sin\alpha-4,698\cdot cos\alpha+12\cdot sin\alpha cos\alpha\\ &M(\alpha)=11,462\cdot (4-x)-4,698y-3\cdot(4-x\cdot \frac{1}{2}(4-x)=\\ &=45,848-45,848\cdot cos\alpha-18,792\cdot sin\alpha-1,5(16-8x+x^2)=\\ &=45,848-45,848\cdot cos\alpha-18,792\cdot sin\alpha-24+48\cdot cos\alpha-24cos^2 \alpha=\\ &21,848+2,152\cdot cos\alpha-18,792\cdot sin\alpha-24\cdot cos^2 \alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[^{\circ}\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 0 & 0.000 & 1.000 & -11.462 & -4.698 & 0.000 \\ \hline 15 & 0.259 & 0.966 & -11.892 & -1.677 & -3.329 \\ \hline 30 & 0.500 & 0.866 & -10.883 & 0.859 & -3.684 \\ \hline 45 & 0.707 & 0.707 & -8.942 & 2.298 & -1.918 \\ \hline 60 & 0.866 & 0.500 & -6.800 & 2.381 & 0.650 \\ \hline 75 & 0.966 & 0.259 & -5.203 & 1.264 & 2.646 \\ \hline 90 & 1.000 & 0.000 & -4.698 & -0.538 & 3.056 \\ \hline \end{array}
Calculation of support reactions
We use the equations of equilibrium of forces and moments to determine the reactions at the supports.
Equations of static equilibrium \begin{aligned} &\sum{x}=0 & H_A-4,698=0 && H_A=4,698 \ kN\\ &\sum{M_A}=0 & 3\cdot 4\cdot 2+1,71\cdot 6-4,698\cdot 2\sqrt3-V_B\cdot 8=0 && V_B=2,248 \ kN\\ &\sum{y}=0 & V_A-3\cdot 4-1,71+2,248=0 && V_A=11,462 \ kN\\ \end{aligned}
Interval I - \(\alpha \in (0;60)\)
We solve for internal forces in the arch in the angular range from 0 to 60 degrees.
Projecting the reaction VB onto the normal and shear components to the mental section of the arch.
We express the variables x and y in terms of the angle alpha.
\begin{aligned} &\frac{x}{4}=cos\alpha\\ &x=4cos\alpha\\ &\frac{y}{4}=sin\alpha\\ &y=4sin\alpha\\ \end{aligned} Internal force functions in the first interval.
\begin{aligned} &N(\alpha)=-2,248 cos\alpha\\ &T(\alpha)=-2,248 sin\alpha\\ &M(\alpha)=2,248 (4-x)=8,992-8,992 cos\alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[{ }^0\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 0 & 0.000 & 1.000 & 2.248 & 0.000 & 0.000 \\ \hline 15 & 0.259 & 0.966 & 2.171 & -0.582 & 0.306 \\ \hline 30 & 0.500 & 0.866 & 1.947 & -1.124 & 1.205 \\ \hline 45 & 0.707 & 0.707 & 1.590 & -1.590 & 2.634 \\ \hline 60 & 0.866 & 0.500 & 1.124 & -1.947 & 4.496 \\ \hline \end{array}
Interval II - \(\alpha \in (60;90)\)
We solve for internal forces in the arch in the angular range from 60 to 90 degrees.
Internal force functions in the second interval.
\begin{aligned} &N(\alpha)=-2,248 cos\alpha+1,71 cos\alpha-4,698 sin\alpha=-0,538 cos\alpha-4,698 sin\alpha\\ &T(\alpha)=-2,248\cdot sin\alpha+1,71\cdot sin\alpha+4,698 cos\alpha=-0,583 sin\alpha+4,698 cos\alpha\\ &M(\alpha)=2,248\cdot (4-x)-1,71\cdot (2-x)-4,698(y-2\sqrt3)=\\ &=8,992-8,992\cdot cos\alpha-3,42+6,84\cdot cos\alpha-18,792 sin\alpha+16,274=\\ &=21,848-2,152 cos\alpha-18,792 sin\alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[^{\circ}\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 60 & 0.866 & 0.500 & -4.338 & 1.883 & 4.496 \\ \hline 75 & 0.966 & 0.259 & -4.677 & 0.696 & 3.139 \\ \hline 90 & 1.000 & 0.000 & -4.698 & -0.538 & 3.056 \\ \hline \end{array}
Interval III - Change of coordinate system \(\alpha \in (0;90)\)
We change the coordinate system, consider the section in the view from the left side, and recalculate the internal forces over the full angular range.
Internal force functions in the third interval.
\begin{aligned} &N(\alpha)=-11,462\cdot cos\alpha -4,698\cdot sin\alpha+3\cdot (4-x) cos\alpha=\\ &=-11.462\cdot cos\alpha-4,698\cdot sin\alpha+12\cdot cos\alpha-12cos^2 \alpha=\\ &=0,538\cdot cos\alpha-4,698\cdot sin\alpha-12\cdot cos^2 \alpha\\ \\ &T(\alpha)=11,462\cdot sin\alpha-4,698\cdot cos\alpha-3\cdot (4-x)\cdot sin\alpha=\\ &=11,462\cdot sin\alpha-4,698\cdot cos\alpha-12\cdot sin\alpha+12\cdot sin\alpha cos\alpha=\\ &=-0,538\cdot sin\alpha-4,698\cdot cos\alpha+12\cdot sin\alpha cos\alpha\\ &M(\alpha)=11,462\cdot (4-x)-4,698y-3\cdot(4-x\cdot \frac{1}{2}(4-x)=\\ &=45,848-45,848\cdot cos\alpha-18,792\cdot sin\alpha-1,5(16-8x+x^2)=\\ &=45,848-45,848\cdot cos\alpha-18,792\cdot sin\alpha-24+48\cdot cos\alpha-24cos^2 \alpha=\\ &21,848+2,152\cdot cos\alpha-18,792\cdot sin\alpha-24\cdot cos^2 \alpha\\ \end{aligned} We check the values of internal forces for the angle alpha every 15 degrees. The results can be conveniently arranged in a table.
\begin{array}{|c|c|c|c|c|c|} \hline \alpha\left[^{\circ}\right] & \sin \alpha & \cos \alpha & \mathrm{N}(\alpha)[\mathrm{kN}] & \mathrm{T}(\alpha)[\mathrm{kN}] & \mathrm{M}(\alpha)[\mathrm{kNm}] \\ \hline 0 & 0.000 & 1.000 & -11.462 & -4.698 & 0.000 \\ \hline 15 & 0.259 & 0.966 & -11.892 & -1.677 & -3.329 \\ \hline 30 & 0.500 & 0.866 & -10.883 & 0.859 & -3.684 \\ \hline 45 & 0.707 & 0.707 & -8.942 & 2.298 & -1.918 \\ \hline 60 & 0.866 & 0.500 & -6.800 & 2.381 & 0.650 \\ \hline 75 & 0.966 & 0.259 & -5.203 & 1.264 & 2.646 \\ \hline 90 & 1.000 & 0.000 & -4.698 & -0.538 & 3.056 \\ \hline \end{array}
Graphs of internal forces in the arch
If you have any questions, comments, or think you have found a mistake in this solution, please send us a message at kontakt@edupanda.pl.