numerical-problems-on-sfd-and-bm

8/7/2019 numerical-problems-on-sfd-and-bm

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Problem: Draw the shear force and bending moment diagrams for the loaded beam.

From the load diagram: usingMD=0, we get30 x 6 + 50 x 2 = RB x 5RB=56kNRD=30 + 50 56 =24 kNConsider a portion of Segment AB at distance x from A

Vx= 30 kN

Mx= 30xkNm

Consider a section in Segment BC at distance x from A

Vx= 30+56, Vx=26 kN

Mx= 30x+56(x1) Mx=26x56 kNm

Consider a section in Segment CD at a distance x from A


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VCD=Vx=30+5650 VCD= 24 kN

MCD= Mx = 30x+56(x1)50(x4) = 30x+56x5650x+200

MCD=24x+144 kNm

To draw the Shear Diagram:

1. In segment AB, the shear is uniformly distributed over the segment at a magnitude of 30 kN.

2. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.3. In segment CD, the shear is uniformly distributed at a magnitude of 24 kN.

To draw the Moment Diagram:

1. The equation MAB = 30x is linear, at x = 0, MAB = 0 and at x = 1 m, MAB = 30 kNm.2. MBC = 26x 56 is also linear. At x = 1 m, MBC = 30 kNm; at x = 4 m, MBC = 48 kNm.When MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B. This point wherebending moment is changing sign is called POINT OF CONTRAFLEXURE.3. MCD = 24x + 144 is again linear. At x = 4 m, MCD = 48 kNm; at x = 6 m, MCD = 0.


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These must be on the same page and one above the other in the order shown.

POINTS OF IMPORTANT OBSERVATION:

(i) Where shear force changes sign are the points of maximum bending moment clear fromSFD and BMD.

(ii) Numerically highest will be maximum shear force from SFD(iii)Numerically highest will be maximum bending moment from BMD


4/21

Problem: Draw the SFD and BMD for the loaded beam shown.

Solution:

MA=0, 10RC=2(80)+5[10(10)], RC=66kN

MC=0, 10RA=8(80)+5[10(10)], RA=114kN

A section in Segment AB:

VAB=11410xkN

MAB=114x10x x/2

MAB=114x5x2 kNm

A section in Segment BC:

VBC=1148010x

VBC=3410x kN

MBC=114x80(x2)10x(x/2)

MBC=160+34x5x2 kNm


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6/21

Problem: Draw the SFD and BMD for the loaded beam.

30 kN/m

Find the reactions by using the equations of equilibrium.

MA=0, 6RD=4[2(30)], RD=40 kN

MD=0, 6RA=2[2(30)], RA=20 kN

Consider segment AB

VAB=20 kN

MAB=20xkNm

Consider segment AC

VBC=2030(x3)

VBC=11030xkN

MBC=20x30(x3)(x3)2

MBC=20x15(x3)2 kNm

Consider segment AD


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VCD=2030(2)

VCD=40 kN

MCD=20x30(2)(x4)

MCD=20x60(x4) kNm


8/21

PROBLEM: Draw the SFD and BMD for the loaded beam.

The maximum bending moment occurs just below where SF changes sign in SFD.


9/21

PROBLEM: DRAW THE SFD AND BMD FOR THE LOADED CANTILEVER BEAM.

Consider segment AB:

VAB=20 kN

MAB=20x kNm

Consider segment AC

VAC=20 kN

MAC=20x+80 kNm


10/21

Problem: Draw the SFD AND BMD for the loaded beams shown.

Problem: Draw the BMD for the loaded beam shown. Also show the actual shape of the bentbeam. Also mark the point of contra-flexure.

Fig : Loaded Beam

Fig. Free Body Diagram


11/21

Fig. Bending moment diagram

Fig. Shape of the bent beam


12/21

PROBLEM:Draw the SFD and BMD for the loaded beam.


13/21

PROBLEM:Draw the shear force and bending moment diagrams.

Fig. BMD

Identify the maximum shear force and bending-moment from SFD and BMD.

Vmax=26 kN and Mmax = 50 kNm


14/21

PROBLEM:A simply supported steel beam is to carry the distributed and concentrated

loads shown. Draw the SFD. From SFD find the location of the maximum bending

moment.

Maximum bending moment occurs where the shear force changes sign i.e. at point E.

Mmax=M2.6 =52 x 2.6 20 x 2.6= 135.252 = 83.2 kNm.


15/21

PROBLEM: Draw the SFD and BMD for the loaded beam with one end hinged and

other end is simply supported.

Four steps: Draw free body diagram. Find reactions by using equations of equilibrium.

Draw SFD. Draw BMD.


16/21

Problem: Draw The SFD and BMD for the loaded beam shown.


17/21

Problem: Draw the SFD and BMD for the loaded beam shown.


18/21

PROBLEM: Draw the SFD and BMD for the loaded beam.


19/21

Problem: Draw the SFD and BMD for the loaded beam shown below. Indicate point ofcontra-flexure if any.


20/21

PROBLEM: Draw the SFD and BMD for the loaded cantilever beam.

HEAR FORCE DIAGRAM AND BENDING MOMENT DIAGRAMS IN THE SHORTEST WAY

Draw loaded beam

Below this draw free body diagram.

Below this draw shear force diagram.

Below this draw bending moment diagram.


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LOADED BEAM, FREE BODY DIAGRAM, SFD and BMD

NOTE: When a concentrated force acts downward then the shear force diagram will jumpdownward at that particular point. When a concentrated force acts upwards then the shear forcediagram will jump upward at that particular point. The UDL will be with an inclined line.When a moment M is applied clockwise on left, the moment diagram will jump upward. When M

acts counterclockwise on left, the moment diagram will jump downward. The points ofconcentrated loads are joined by straight lines and UDL by smooth curves.

numerical-problems-on-sfd-and-bm

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