SFD and BMD for eccentrically point loaded beam

Draw SFD and BMD for a simply supported eccentrically point loaded beam having length L. A point load W is acting at distance a from support A and remaining distance is b.

Solution:

Step 1: Calculate support reactions

$\sum F_{y}$	$=0$
$Upward\: forces$	$= Down\: forces$
$R_{A}+R_{B}$	$=W$ _____Eq. 1
$\sum M_{a}$	$=0$
$+\left (R_{B}\times L \right )-\left ( W\times a \right )$	$=0$
$\left (R_{B}\times L \right )$	$=\left ( W\times a \right )$

$R_{B}=\frac{Wa}{L}$

Put this value in Eq. 1

$R_{A}+\frac{Wa}{L}$	$=W$
$R_{A}$	$=W-\frac{Wa}{L}$

$R_{A}=\frac{W(L-a)}{L}$

Step 2: Calculate Shear forces

1) Shear force just left of point A

There is no any force to left side of point A therefore,

$SF_{LA}=0$

2) Shear force just right of point A

There is only reaction due to support A present in left side of section X-X and just right of point A. Reaction due to support A is upward it taken as +ve because of by sign conventions upward forces left side of section are taken as +ve. Therefore,

$SF_{RA}=+R_{A}$

$SF_{RA}=+\frac{W(L-a)}{L}$

3) Shear force just left of point B

There is only reaction due to support A present in left side of section X-X and just left of point B. Reaction due to support A is upward it taken as +ve because of by sign conventions upward forces left side of section are taken as +ve. Therefore,

$SF_{LB}=+R_{A}$

$SF_{LB}=+\frac{W(L-a)}{L}$

4) Shear force just right of point B

There is reaction due to support A present and point load at C in left side of section X-X and just right of point B. Reaction due to support A is upward it taken as +ve and point load is downward it taken as -ve because of by sign conventions upward forces left side of section are taken as +ve and downward forces are -ve. Therefore,

$SF_{RB}=+R_{A}-W$
$SF_{RB}=+\frac{W(L-a)}{L}-W$
$SF_{RB}=\frac{W(L-a)-WL}{L}$
$SF_{RB}=\frac{WL-Wa-WL}{L}$

$SF_{RB}=-\frac{Wa}{L}$

5) Shear force just left of point C

There is reaction due to support A and point load at C is present in left side of section X-X and just left of point C. Reaction due to support A is upward it taken as +ve and point load is downward it taken as -ve because of by sign conventions upward forces left side of section are taken as +ve and downward forces are -ve. Therefore,

$SF_{LC}=+R_{A}-W$
$SF_{LC}=+\frac{W(L-a)}{L}-W$
$SF_{LC}=\frac{W(L-a)-WL}{L}$
$SF_{LC}=\frac{WL-Wa-WL}{L}$

$SF_{LC}=-\frac{Wa}{L}$

6) Shear force just right of point C

There is reaction due to support A, point load at C and reaction due to support B is present in left side of section X-X and just right of point C. Reaction due to support A is upward it taken as +ve, point load at C is downward it taken as -ve and reaction due to support B is upward it is taken as +ve; because of by sign conventions upward forces left side of section are taken as +ve and downward forces are -ve. Therefore,

$SF_{RC}=+R_{A}-W+R_{B}$
$SF_{RC}=+\frac{W(L-a)}{L}-W+\frac{Wa)}{L}$
$SF_{RC}=\frac{W(L-a)-WL+Wa}{L}$
$SF_{RC}=\frac{WL-Wa-WL+Wa}{L}$

$SF_{RC}=0$

All shear forces are summarized as below;
$SF_{LA}=0$
$SF_{RA}=+\frac{W(L-a)}{L}$
$SF_{LB}=+\frac{W(L-a)}{L}$
$SF_{RB}=-\frac{Wa}{L}$
$SF_{LC}=-\frac{Wa}{L}$
$SF_{RC}=0$

Step 3: Draw Shear forces diagram (SFD)

Now, SFD is drawn as below,
(Note: SFD must be drawn below the beam)

Step 4: Calculate bending moments

1) Bending moment at point A

There is no any force at left of point A. Therefore

$M_{A}=0$

2) Bending moment at point B

At this point only reaction due to support A is present at left side of point B. This moment due to R_A is Clock wise acting at dist. 'a' from 'B'. By sign convention clock wise moment at left side of point is +ve; hence moment due to R_A is +ve. There fore,

$M_{B}=+R_{A}\times Perpendicular\: Dist.\: between\: A\: and\: B$

$M_{B}=+\frac{W(L-a)}{L}\times a$

$M_{B}=+\frac{Wa(L-a)}{L}$

3) Bending moment at point C

At this point there are two forces acting at left side of point C.

1. Reaction due to support A.

2. Point load at B.

1. Reaction due to support A : Reaction due to support A is clock wise and it is taken as +ve because of by sign convention clock wise moment at left side of point is taken as +ve. It acting at 'L' distance form point 'C'.

2. Point load at B : Point load at C is anti-clock wise and it is taken as -ve because of by sign convention anti-clock wise moment at left side of point is taken as -ve. It acting at 'b' distance form point 'C'.
Therefor,
$M_{C}=+\left (R_{A}\times Distance\: bet.\: A\: and\: C \right )-\left ( W\times Distance\: bet.\: B\: and\: C \right )$
$M_{C}=+\left (\frac{W(L-a)}{L}\times L \right )-\left ( W\times b \right )$
$M_{C}=+\left (\frac{WL(L-a)-WLb}{L} \right )$
$M_{C}=+\left (\frac{WL(L-a-b)}{L} \right )$
$M_{C}=+\left (\frac{WL(L-[a+b])}{L} \right )$
$M_{C}=+\left (\frac{WL(L-L)}{L} \right )$
$M_{C}=+\left (\frac{WL\times0}{L} \right )$

$M_{C}=0$