5.7 and 5.8 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment. And 5.10 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

General guidance

Concepts and reason

Equilibrium of a rigid body:

An object is said to be in equilibrium when the sum of external forces and couples acting in the object are zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external couples about any point should be zero.

For a rigid body to be in equilibrium in three dimensions, the sum of external forces acting along , and directions have to zero.

Shear force:

When force acting on a body pushes two parts of the body in opposite direction then the force is referred as shear force. Shear force on an object acts normal to its cross section.

Bending moment:

It is the degree of bending caused on the beam when an arbitrary load acts on it. Usually moment is said to prevail on a beam or structure that is acted upon by a perpendicular force triggering rotation.

The shear force values at different points of the beam can be calculated using force equilibrium condition. Similarly the bending moment values at different points of the beam can be calculated using moment equilibrium condition.

Fundamentals

Write the equilibrium of forces along the -axis.

Here, the sum of forces along the x-direction is .

Write the equilibrium of forces along the -axis.

Here, the sum of forces along the y-direction is .

Write the relation for equilibrium of moments.

Here, the sum of moments about any point is .

Sign convention of force:

The forces acting towards right side are considered as positive and vice versa. Similarly upward vertical forces are considered as positive and vice versa.

Sign convention of moment:

The moments which rotate the body in counter clockwise direction are taken as positive and the moments which rotate the body in clockwise direction are taken as negative.

Step-by-step

Step 1 of 13

5.7)

Draw the free body diagram of the beam.

Apply the force equilibrium condition and balance all vertical forces.

Take moments about the point A.

Calculate the sum of forces acting along the -axis.

The free body diagram of beam is drawn by replacing the fixed support at A with reaction forces and moment. The reaction moment at support A is calculated by applying moment equilibrium condition about point A. then the components of reaction force acting at support A are calculated by applying force equilibrium condition.

The forces acting along the vertical direction should not be considered while calculating the sum of forces acting along the x-axis.

Calculate the shear force values at various locations of the beam and use the obtained values to draw the shear force diagram.

Step 2 of 13

Calculate the Shear force at the point A as shown:

Calculate the Shear force at the point C as shown:

Calculate the Shear force at the point left of D as shown:

Calculate the Shear force at the point B as shown:

The shear force values at different locations of the beam are calculated.

Calculate the bending moment values at various locations of the beam and use the obtained values to draw the bending moment diagram.

Step 3 of 13

Calculate the Bending moment at the point A.

Calculate the Bending moment at the point C as shown:

Calculate the Bending moment at the point D as shown:

Calculate the Bending moment at the point E as shown:

Calculate the Bending moment at the point B as shown:

Draw the shear force and bending moment diagram by plotting the beam length along the x-axis and magnitudes of the shear force and bending moments along the y-axis. Use different locations of the beam and corresponding shear force and bending moment values to draw a complete shear force and bending moment diagram.

Draw the shear force diagram and bending moment diagram.

Part 5.7

The shear force diagram for the beam is as below:

The shear force and bending moment values at different locations of the beam are calculated. The shear force diagram and bending moment diagrams are is drawn by plotting the beam length along the x-axis and magnitudes of the shear force and bending moments along the y-axis.

Calculate the maximum shear force.

Step 4 of 13

5.7.a)

Calculate the maximum shear force using the shear force and bending moment diagram.

Part 5.7.a

The maximum shear force is

The maximum shear force is calculated by using the shear force diagram.

Calculate the maximum bending moment.

Step 5 of 13

5.7.b)

Calculate the maximum shear force using the shear force and bending moment diagram.

Part 5.7.b

The maximum bending moment is .

The maximum bending moment is calculated by using the bending moment diagram.

Use the equilibrium relations of forces and moments to calculate the reaction forces.

Step 6 of 13

5.8)

Draw the free body diagram of the beam.

Consider moment equilibrium equation about point C.

Here, is the vertical reaction force at point E.

Apply vertical force equilibrium equation as follows:

Here, is the vertical reaction force at point C.

Substitute for .

Apply horizontal force equilibrium equation as follows:

Here, is the horizontal reaction force at point C.

The free body diagram of beam is drawn by replacing the pin support at C with reaction forces. The reaction forces at support E is calculated by applying moment equilibrium condition about point C. then the components of reaction force acting at support C are calculated by applying force equilibrium condition.

The forces acting along the vertical direction should not be considered while calculating the sum of forces acting along the x-axis.

Calculate the shear force values at various locations of the beam and use the obtained values to draw the shear force diagram.

Step 7 of 13

Consider the section between A and C.

Apply vertical force equilibrium equation.

Calculate the shear force at the point A.

At ,

Calculate the shear force at the point left of C.

At ,

Consider the section between C and D.

Calculate the shear force at the point right of C.

Substitute for and for .

Apply vertical force equilibrium equation.

Substitute for

Calculate the shear force at the point left of point D.

At ,

The shear force values at different locations of the beam are calculated. The shear force diagram is drawn by plotting the beam length along the x-axis and magnitudes of the shear force along the y-axis.

Calculate the bending moment values at various locations of the beam and use the obtained values to draw the bending moment diagram.

Step 8 of 13

Calculate the Bending moment at the point A.

Calculate the Bending moment at the point C as shown:

Calculate the Bending moment at the point D as shown:

Calculate the Bending moment at the point E as shown:

Calculate the Bending moment at the point B as shown:

Draws the shear and bending moment diagrams for the beam and loading as follows:

Part 5.8

The shear force and bending moment diagram for the beam is as below:

The shear force and bending moment values at different locations of the beam are calculated. The shear force and bending moment diagram is drawn by plotting the beam length along the x-axis and magnitude of the shear force and bending moment along the y-axis.

Calculate the maximum shear force.

Step 9 of 13

5.8.a)

Calculate the maximum shear force using the shear force and bending moment diagram.

Part 5.8.a

The maximum shear force is

The maximum shear force is calculated by using the shear force diagram.

Calculate the maximum bending moment.

Step 10 of 13

5.8.b)

Calculate the maximum shear force using the shear force and bending moment diagram.

Part 5.8.b

The maximum bending moment is .

The maximum bending moment is calculated by using the bending moment diagram.

Draw the free body diagram of the beam.

Step 11 of 13

Draw the free body diagram of the beam.

Here, is the horizontal reaction force at point , is the vertical reaction force at point , and is the vertical reaction force at point .

Note: Since the support at point is a roller support horizontal component of the reaction force at this point is zero.

Determine the magnitude of the reaction forces.

Take moments about point .

Take moments about point .

Consider the equilibrium of the forces along the horizontal direction:

Draw the free body diagram and using the equilibrium equations to be calculated reaction forces.

Determine the equations for the shear force and the bending moment.

5.1.b)

Step 12 of 13

Determine the equations for the shear force and the bending moment.

Take an arbitrary section of the beam of length from point such that and consider the equilibrium of the section.

Here, is the shear force at point , and is the bending moment at point .

Determine the equation for the shear force for the following range: .

Consider the equilibrium of the forces in the vertical direction.

…… (1)

From equation (1) it is clear that the shear force curve is linear.

Determine the equation for the bending moment for the following range: .

Take moments about point .

…… (2)

From equation (2) it is clear that the bending moment curve is parabola.

Part 5.1.b

The shear force and bending moment equations are and .

Take section and using the equilibrium equations to be calculated shear force and bending moment equations.

Calculate the shear force values at various locations of the beam and use the obtained values to draw the shear force diagram.

5.1a)

Step 13 of 13

Calculate the shear force at point .

Substitute for in the equation (1).

Calculate the shear force at the midpoint of the beam.

Substitute for in the equation (1).

Calculate the shear force at point .

Substitute for in the equation (1).

Calculate the bending moment at point .

Substitute for in the equation (2).

Calculate the bending moment at the midpoint of the beam.

Substitute for in the equation (2).

Calculate the bending moment at point .

Substitute for in the equation (2).

Draw the Shear force and bending moment diagrams using the magnitudes of the shear force and the bending moment at the salient points and the equations (1) and (2).

Part 5.1.a

The shear force and bending moment diagram for the beam is as below:

The shear force and bending moment values at different locations of the beam are calculated. The shear force and bending moment diagram is drawn by plotting the beam length along the x-axis and magnitude of the shear force and bending moment along the y-axis.

Answer

Part 5.7

The shear force diagram for the beam is as below:

Part 5.7.a

The maximum shear force is

Part 5.7.b

The maximum bending moment is .

Part 5.8

The shear force and bending moment diagram for the beam is as below:

Part 5.8.a

The maximum shear force is

Part 5.8.b

The maximum bending moment is .

Part 5.1.b

The shear force and bending moment equations are and .

Part 5.1.a

The shear force and bending moment diagram for the beam is as below:

Answer only

Part 5.7

The shear force diagram for the beam is as below:

Part 5.7.a

The maximum shear force is

Part 5.7.b

The maximum bending moment is .

Part 5.8

The shear force and bending moment diagram for the beam is as below:

Part 5.8.a

The maximum shear force is

Part 5.8.b

The maximum bending moment is .

Part 5.1.b

The shear force and bending moment equations are and .

Part 5.1.a

The shear force and bending moment diagram for the beam is as below:

ΣF = 0

ΣΕ, = 0

ΣΜ = 0

3 kN 2 KN 5 kN 2 KN ΜΑ ET 0.3 m 2 0.3 m 0.4

ΣΕ, = 0 Α, +5=3+2+2 Α = 2kΝ

ΣΜ, = 0 Μ, = (2) (0.3 + 0.3 + 0.3 + 0.4)-(5)(0.3 + 0.3 + 0.3) + (2) (0.3 + 0.3) + (3)(0.3) = 0.2kΝ – m

ΣΕ = 0 Α = 0

Vi= 4, = 2 kN

Vc = 4, -3 = 2-3 =-1kN

V = 4,-3-2 = 2-3-2 =-3kN

Ve = 4,-3-2+5 = 2-3-2+5 = 2 kN

Vo = 4, -3-2+5-2 = 0kN

BM =-M = -0.2 kN.m

BM, 5-M, +(4,0.3) =-(0.2)+(2×0.3) = 0.4 kN.m

BM, 5-M, +(4×0.6)-(3×0.3) =-0.2+(2×0.6)-(3×0.3) = 0.1kN.m

BM, =-M,+(4,0.9)-(3×0.6)-(2×0.3) =-0.2+(2×0.9)-(3×0.6)-(2×0.3) = -0.8kN.m

BM3 =-M,+(4,-1.3)-(3×1.0)-(2×0.7)+(5×0.4) =-0.2+(2×1.3)-(3×1.0)-(2×0.7)+(5×0.4) = OKN.m

3 kN KN 5 kN 2 C D E A M = 0.2 kN m L RA = 2 kNK 0.3 m 0.3 m 0.3 m 0.4 m 2 kN 2 kN – 1 kN Shear force diagram – 3 kN 0.4 kN m 0.1kN m 0 kN m 7 CDT – 0.2 kN m Bending moment diagram – 0.8 kN m

SFmax = 3kN

BM = 0.8kN.m

100 lb 250 lb 100 lb AC AC D 15 in. 1 20 in. I 25 in 25 in. 10 in

ΣΜ. = 0 – (100)(20+25 +10)+(Ε,)(20+25)-(250)(20) + (100) (15) = 0 -5500 + (Ε, x45)-5000 +1500 = 0 (Ε, x45) = 9000 Ε, = 200 16

ΣF, = 0 C, +Ε = 100 – 250 +100

200 lb

C = 450-200 = 250 lb

ΣF = 0 C, = 0

100 lb

ΣΕ = 0 V+100 = 0 V=-10OIb

Vi =-100 lb

x=15in

Vy=-1001b

100 lb M 15 in

Vc, =Vc/+C,

250 lb

-100 lb

ΣΕ, = 0 V+100-C, = 0 V=C, -100

250 lb

V = 250-100 = 150 lb

x= 35 in

Vp , = 150 lb

BM =-M = -0.2 kN.m

BM, 5-M, +(4,0.3) =-(0.2)+(2×0.3) = 0.4 kN.m

BM, 5-M, +(4×0.6)-(3×0.3) =-0.2+(2×0.6)-(3×0.3) = 0.1kN.m

BM, =-M,+(4,0.9)-(3×0.6)-(2×0.3) =-0.2+(2×0.9)-(3×0.6)-(2×0.3) = -0.8kN.m

BM3 =-M,+(4,-1.3)-(3×1.0)-(2×0.7)+(5×0.4) =-0.2+(2×1.3)-(3×1.0)-(2×0.7)+(5×0.4) = OKN.m

100 lb 250 lb 100 lb A C D 15 in. I 20 in. . 25 in. 10 in. 1001b AI 1001b 1001b 1500 lb-in M TO -1000 lb in -1500 lb-in

SFmax = 150 lb

BM = 15001b-in

ΣF = 0 Α, = 0

(= (20xм охтм )

3 KN 2 KN SKNI 2011 DE M, = 0.2 kN m R=2 nk 0.3 m 0.3 m 0.3 m 0.4 m 2 KN 2 KN D – 1 kN Shear force diagram – 3 KN 0.4 kNm 0.1kN m 0 kNm C D E -0.2 kN m Bending moment diagram -0.8 kNm

3 KN 2 KN SKNI 2011 DE M, = 0.2 kN m R=2 nk 0.3 m 0.3 m 0.3 m 0.4 m 2 KN 2 KN D – 1 kN Shear force diagram – 3 KN 0.4 kNm 0.1kN m 0 kNm C D E -0.2 kN m Bending moment diagram -0.8 kNm

3 kN 2 KN 5 kN 2 ķN [ 0.3 m D 1 0.3 m E 0.3 m 0.4 m. A, IkN 3kN Nm DIE

0.8kNôm

0.8kNôm

100 lb 250 lb 100 lb A C D 15 in. I 20 in. . 25 in. 10 in. 1001b AI 1001b 1001b 1500 lb-in M TO -1000 lb in -1500 lb-in

100 lb 250 lb 100 lb A C D 15 in. I 20 in. . 25 in. 10 in. 1001b AI 1001b 1001b 1500 lb-in M TO -1000 lb in -1500 lb-in

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