Chapter 4, Problem 4.1P

To determine

The internal normal force, shear force and bending moment at C and D.

Answer to Problem 4.1P

$\begin{array}{l}{\text{Internal normal force at C, N}}_{\text{C}}=0\\ {\text{Shear force at C, V}}_{\text{C}}=-2.5\text{ kN}\\ {\text{Bending moment at C, M}}_{\text{C}}=-35\text{ kNm}\\ {\text{Internal normal force at D, N}}_{\text{D}}=0\\ {\text{Shear force at D, V}}_{\text{D}}=-20\text{ kN}\\ {\text{Bending moment at D, M}}_{\text{D}}=-40\text{ kNm}\end{array}$

Explanation of Solution

Given Information:

Point D is located just right to the concentrated load. Support B is a pin and A is a roller.

Method of sections is used to find the values.

Calculation:

Initially the reactions at supports are to be found.

Moment about A

$\begin{array}{l}{\displaystyle \sum {\text{M}}_{\text{A}}}=0\\ (30\times 1)+(15\times 4)+30-({\text{R}}_{\text{B}}\times 6)=0\\ {\text{R}}_{\text{B}}=20\text{ kN}\\ {\text{R}}_{\text{A}}+{\text{R}}_{\text{B}}=30+15+(\frac{30}{4})\\ {\text{R}}_{\text{A}}=32.5\text{ kN}\end{array}$

Internal Normal Force, Shear force and Bending moment at C:

By using method of section in the section AC

$\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{N}}_{\text{C}}=0\\ {\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{R}}_{\text{A}}+{\text{V}}_{\text{C}}=15\times 2\\ {\text{V}}_{\text{C}}=-2.5\text{ kN}\\ {\displaystyle \sum {\text{M}}_{\text{A}}}=0\\ (30\times 1)-(-2.5\times 2)+{\text{M}}_{\text{C}}=0\\ {\text{M}}_{\text{C}}=-35\text{ kNm}\end{array}$

Internal Normal Force, Shear force and Bending moment at D:

By using method of section in the section DB

$\begin{array}{l}{\displaystyle \sum {\text{F}}_{\text{x}}}=0\\ {\text{N}}_{\text{D}}={\text{H}}_{\text{B}}=0\\ {\displaystyle \sum {\text{F}}_{\text{y}}}=0\\ {\text{R}}_{\text{B}}+{\text{V}}_{\text{D}}=0\\ {\text{V}}_{\text{D}}=-20\text{ kN}\\ {\displaystyle \sum {\text{M}}_{\text{B}}}=0\\ -(-20\times 2)+{\text{M}}_{\text{D}}=0\\ {\text{M}}_{\text{D}}=-40\text{ kNm}\end{array}$

Conclusion:

Therefore,

$\begin{array}{l}{\text{Internal normal force at C, N}}_{\text{C}}=0\\ {\text{Shear force at C, V}}_{\text{C}}=-2.5\text{ kN}\\ {\text{Bending moment at C, M}}_{\text{C}}=-35\text{ kNm}\\ {\text{Internal normal force at D, N}}_{\text{D}}=0\\ {\text{Shear force at D, V}}_{\text{D}}=-20\text{ kN}\\ {\text{Bending moment at D, M}}_{\text{D}}=-40\text{ kNm}\end{array}$