Subject: Structural Analysis 1

Topic: Axial Force, Shear Force and Bending Moment

Difficulty: High

$\begin{align} &\text{ B is an internal hinge, B.M at B = 0 } \therefore \text{ considering BEFC,} \\ & BM_B=0, \ \text{gives, } \\& \phantom{BM_B=0, \ \text{gives, }} V_C\times5-8\times2-5\times4=0 \\& \phantom{BM_B=0, \ \text{gives, }}\boxed{ V_C=7.2 \ kN} {\uparrow} \\ & \sum F_Y=0 \ (\uparrow +ve) \\ & V_A+7.2-8-5=0 \\ & \boxed{V_A = 5.8\ kN} \uparrow \\ \\& \sum F_Y=0 \ (\rightarrow +ve) \\ & -H_A+10+5\times4=0 \\& \boxed{ H_A=30 \ kN} \leftarrow \\& \text { Assuming an anticlockwise resting moment } M_A \ at \ a, \text{as shown} \\&\text{consider part BA; B.M at A=0 }[\circlearrowright | \circlearrowleft +ve \ \circlearrowleft|\circlearrowright -ve] \\& 30\times4-M_A-10\times2-5\times4\times2=0 \\& \boxed{M_A=60\ kNm} \\& \text{Answer for } M_A\text{ came out positive, so assumption of direction of } M_A\text{ is correct.} \end{align}$

1.Consider member ADB

SF Calculations:

$\begin{align} &S.F_A=+30 \ kN \\ & S.F_{D(below)}=+30-10=20 \ kN \\& S.F_{D(above)}=+20-10 =10 \ kN \\& S.F_B=0 \\ \end{align}$

Bending Moment calculations:

$\begin{align} &B.M_A=-60\ kNm \ \ \ \{trying \ to\ bend\ A\ inside\} \\&B.M_D=30\times2-60-5\times2\times1+5.8\times0\\ &\phantom{B.M_D}= -10\ kNm \\&B.M_D=0 \ \ \ \{ \therefore Internal \ hinge \} \end{align}$

2.Consider member BC:

$\begin{align} &\textbf{SF calculations: } [\uparrow | \downarrow +ve] \\ &S.F_{B(L)}=0 \\ &S.F_{B(R)}=5.8 \ kN \\ &S.F_{E(L)}=5.8 \ kN \\ &S.F_{E(R)}=5.8-8=-2.2 \ kN \\ &S.F_{F(L)}=-2.2 \ kN \\ &S.F_{F(R)}=-7.2 \ kN \\ &S.F_{C}=-7.2 \ kN \end{align}$

$\begin{align} &\textbf{BM calculations: } [\circlearrowright | \circlearrowleft +ve] \\ &BM_{B}=0 \ \ \ \ \ \ \text{\{Internal hinge\} } \\ &BM_{E}=7.2\times3-5\times2=\underline {11.6 \ kNm} \\ &BM_{F}=7.2\times1=\underline {7.2\ kNm} \\ &BM_{C}=0 \ \ \ \ \ \ \text{\{Roller support\} } \end{align}$