Lecture 4.pdffor cicil engineering students

MECHANICS OF SOLIDS-I
MODULE-4
SF AND BM DIAGRAMS OF BEAMS
PROF. DR. MOHAMMAD ASHRAF
DEPARTMENT OF CIVIL ENGINEERING, UET PESHAWAR

CONTENTS
• Introduction
• Shear Force and Bending Moment in Beams
• Shear force and Bending Moment Diagrams
• By Equation Method
• By Area Method
• Examples and Problems
2

INTRODUCTION
Beam is a structure or a structural member which carry transverse loadings (loads
perpendicular to the longitudinal axis of the member).
Consider a beam along x-axis is subjected to transverse loading along the y-axis. The transverse
loading on beam induces shear forces parallel y-axis and bending moments about z-axis.
3

SHEAR FORCE IN BEAMS
Shear Force:
Shear force at a certain section in a beam is defined as sum of all forces to the left or right
of that section.
For a beam in equilibrium, shear force at a certain section calculated as sum of forces to the
left will be equal to shear force calculated as sum of forces to the right of that section.
If at a certain section a concentrated force is acting, then there will be two shear values at
that section, i.e. shear force just to the left and shear force just to the right of that section.
Sign Convention: If the tendency of shear force is to slide left portion up with respect to
right portion, it is said to be positive shear force.
If the shear force is calculated from left side, upward forces are taken as positive and
downward forces are taken as negative.
If the shear force is calculated from right side, downward forces are taken as positive and
upward forces are taken as negative.
4

SHEAR FORCE IN BEAMS (CONT..)
Shear Force Example: For the given beam at section 1-1,
Shear calculated from left side,
V1 = 7.5 + (-2x7.5) = -7.5 k
Shear calculated from right side,
V1 = +10 + (-32.5) + (2x7.5) = -7.5 k
So it is verified that shear force calculated from left side is same as shear force calculated from
right side.
At point D, there will be two values of shear force as there is concentrated force at this point.
VD, Left = 7.5 + (-2x15) = -22.5 k (section just to the left of D)
VD, Right = 7.5 + (-2x15) + 32.5 = 10 k (section just to the right of D)
5

BENDING MOMENT IN BEAMS
Bending Moment:
Bending moment at a certain section in a beam is defined as sum of all moments to the left or right of
that section.
For a beam in equilibrium, bending moment at a certain section calculated as sum of moments to the
left will be equal to bending moment calculated as sum of moments to the right of that section.
If at a certain section a concentrated moment is acting, then there will be two bending moment values
at that section, i.e. bending moment just to the left and bending moment just to the right of that
section.
Sign Convention: A bending moment is said to be positive if it cause tensile stresses in bottom fibers
and compressive stress in top fibers.
If the bending moment is calculated from left side, clockwise moments are taken as positive and
counter clockwise moments are taken as negative.
And if the bending moment is calculated from right side, counter clockwise moments are taken as
positive and clockwise moments are taken as negative.
6

BENDING MOMENT IN BEAMS (CONT..)
Bending Moment Example: For the given beam at section 1-1,
Bending moment calculated from left side,
M1 = 7.5 x 22.5 + (-100) + (-2 x 7.5 x 7.5/2)
= +12.5 k-ft
Bending moment calculated from right side,
M1 = (-10 x 17.5) + (32.5 x 7.5) + (-2 x 7.5 x 7.5/2) = +12.5 k-ft
So both resulted in same value.
At point B, there will be two values of bending moment as there is concentrated moment at this
point.
MB, Left = +7.5 x 7.5 = +56.25 k-ft (section just to the left of B)
MB, Right = +7.5 x 7.5 + (-100) = -43.75 k-ft (section just to the right of B)
7

S.F. AND B.M. DIAGRAMS
Diagrams showing variations of shear forces and bending moments along the length of a beam
are known respectively as shear force and bending moment diagrams.
There are two methods of developing shear force and bending moment diagrams:
1. By developing shear force and bending moment functions for various segments of the
beams.
2. By area method in which the shear force diagram is drawn from load diagram area and
bending diagram is drawn from shear force diagram area.
8

S.F. AND B.M. FUNCTIONS
The variation of shear force and bending moment along the length of beam, written as a functions of
position (x) of an arbitrary section, are known respectively as shear force and bending functions.
Different functions are written for different segments of the beam separated by change points. The
change point is:
1. Where concentrated force or moment is acting,
2. Where distributed load starts or ends or changes its magnitude or type.
The position (x) of various functions may refer to same or different origins. However, the applicable
range of (x) shall correspond to start and end of a particular segment with respect to the selected
origin, e.g.:
For AB, 0 ≤ x1 ≤ 7.5
For BC 7.5 ≤ x2 ≤ 15
For CD 15 ≤ x3 ≤ 30
For DE 0 ≤ x4 ≤ 10
9

S.F. AND B.M. FUNCTIONS (CONT..)
Example: Shear Force Functions:
For AB, V1 = 7.5 0 ≤ x1 ≤ 7.5
For BC, V2 = 7.5 7.5 ≤ x2 ≤ 15
For CD, V3 = 7.5 – 2.0 (x3 – 15) = 7.5 – 2x3 + 30
= 37.5 – 2x3 15 ≤ x3 ≤ 30
For DE, V4 = 10 0 ≤ x4 ≤ 10
10

Example: Bending Moment Functions:
For AB, M1 = 7.5 x1 0 ≤ x1 ≤ 7.5
For BC, M2 = 7.5 x2 – 100 7.5 ≤ x2 ≤ 15
For CD, M3 = 7.5 x3 – 100 – 2.0 (x3 – 15) (x3 – 15) /2 = 7.5 x3 – 100 – (x3 – 15)2
= 7.5 x3 – 100 – (x3
2 – 30 x3 + 225) = – x3
2 + 37.5 x3 – 325 15 ≤ x3 ≤ 30
For DE, M4 = -10 x4 0 ≤ x4 ≤ 10
11

Example: Shear Force Diagram:
For AB, V1 = +7.5 0 ≤ x1 ≤ 7.5
At x1 = 0, VA = +7.5 k, At x1 = 7.5, VB = +7.5 k
For BC, V2 = +7.5 7.5 ≤ x2 ≤ 15
At x2 = 7.5, VB = +7.5 k, At x2 = 15, VC = +7.5 k
For CD, V3 = +37.5 – 2x3 15 ≤ x3 ≤ 30
At x3 = 15, VC = +37.5 – 2 (15) = +7.5 k
At x3 = 30, VD = +37.5 – 2 (30) = -22.5 k
The point of zero shear can be calculated from:
V3 = 0 => 37.5 – 2x3 = 0 => x3 = 18.75 ft
For DE, V4 = +10 0 ≤ x4 ≤ 10
At x4 = 10, VD = +10 k, At x4 = 0, VE = +10 k
12

Example: Bending Moment Diagram:
For AB, M1 = 7.5 x1 0 ≤ x1 ≤ 7.5
At x1 = 0, MA = 7.5 (0) = 0 At x1 = 7.5, MB,Left = 7.5 (7.5) = +56.25 k-ft
For BC, M2 = 7.5 x2 – 100 7.5 ≤ x2 ≤ 15
At x2 = 7.5, MB, Right = = 7.5 (7.5) – 100 = -43.75 k-ft
At x2 = 15, MC = 7.5 (15) – 100 = +12.5 k-ft
For CD, M3 = – x3
2 + 37.5 x3 – 325 15 ≤ x3 ≤ 30
At x3 = 15, MC = – (15) 2 + 37.5(15) – 325 = +12.5 k-ft
At x3 = 18.75, Mmax = – (18.75) 2 + 37.5(18.75) – 325 = +26.5625 k-ft
At x3 = 30, MD = – (30) 2 + 37.5(30) – 325 = -100 k-ft
For point of zero moment, M3 = 0 => – x3
2 + 37.5 x3 – 325 = 0 => x3 = 23.904 ft
For DE, M4 = -10 x4 0 ≤ x4 ≤ 10
At x4 = 10, MD = -10 (10) = -100 k-ft, At x4 = 0, ME = -10 (0) = 0
13

SF AND BM DIAGRAMS
Example 2: Analyze the given determinate beam, develop shear force and bending functions and
draw shear force and bending diagrams.
Solution:
Step No.1: Determine support reactions using equilibrium equations.
ΣMB, Left = 0 => VA x 10 + 60 – (2 x 10) (10/2) = 0 => VA = 4.0 k
ΣMF = 0 => 4.0 x 32 + 60 - (2 x 10) (27) -5.0 (16) + VD x 12 – (3 x 12 / 2) (6) = 0 => VD = 45 k
ΣFY = 0 => 4.0 + 45 + VF –(2 x 10) – 5.0 – (3 x 12 / 2) = 0 => VF = -6 k
14

SF AND BM DIAGRAMS
Example 2 (Cont..):
Step No.2: Develop shear force functions for various segments.
V1 = 4.0 – 2x1 = -2x1 + 4.0 0 ≤ x1 ≤ 10
V2 = 4.0 – (2 x 10) = -16.0 10 ≤ x2 ≤ 16
V3 = 4.0 – (2 x 10) – 5.0 = -21.0 16 ≤ x3 ≤ 20
V4 = 4.0 – (2 x 10) – 5.0 + 45 - (12 – x4) y/2 = +24 - (12 – x4)2 / 4 6 ≤ x4 ≤ 12
V5 = +6.0 + x5 y /2 = +6.0 + x5
2 / 4 0 ≤ x5 ≤ 6
15
y /3 = (12-x4) /6
=> y = (12-x4) /2
y /3 = x5 /6
=> y = x5 /2

SF AND BM DIAGRAMS
Example 2 (Cont..):
Step No.3: Develop bending moment functions for various segments.
M1 = 60 + 4 (x1) – 2x1(x1/2) = 60 + 4x1 - x1
2 0 ≤ x1 ≤ 10
M2 = 60 + 4 (x2) – (2 x 10)(x2-5) = 160 – 16x2 10 ≤ x2 ≤ 16
M3 = 60 + 4 (x3) – (2 x 10)(x3-5) – 5 (x3 – 16) = 240 – 21x3 16 ≤ x3 ≤ 20
M4 = 60 + 4 (32-x4) – (2 x 10)(27-x4) – 5 (16-x4) +45 (12-x4) – ((12-x4)2/4) ((12-x4)/3)
M4 = 108 – 24x4 - (12-x4)3/12 6 ≤ x4 ≤ 12
M5 = -6.0 x5 - (x5
2 /4)(x5 / 3) = -6.0 x5 - x5
3 / 12 0 ≤ x5 ≤ 6
16

SF AND BM DIAGRAMS
Example 2 (Cont..):
Step No.4: Draw shear force diagram.
V1 = -2x1 + 4.0 => VA = -2(0) + 4.0 = 4.0 k
VB = -2(10) + 4.0 = -16.0 k
Point of zero shear, -2x1 + 4.0 = 0 => x1 = 2.0 ft
V2 = -16.0 k => VC, Left = -16.0 k
V3 = -21.0 k => VC, Right = -21.0 k
VD, Left = -21.0 k
V4 = +24 - (12 – x4)2 / 4
VD, Right = +24 - (12 – 12)2 / 4 = +24.0 k
VE = +24 - (12 – 6)2 / 4 = +15.0 k
V5 = +6.0 + x5
2 / 4 => VF = +6.0 + (0)2 / 4 = +6.0 k
17

SF AND BM DIAGRAMS
Example 2 (Cont..):
Step No.4: bending moment diagram.
M1 = 60 + 4x1 - x1
2 => MA = 60 + 4(0) – (0)2 = +60 k-ft
MV=0 = 60 + 4(2) – (2)2 = +64 k-ft
MB = 60 + 4(10) – (10)2 = 0
M2 = 160 – 16x2=>MC = 160 – 16(16) = -96 k-ft
M3 = 240 – 21x3 => MD = 240 – 21(20) = -180 k-ft
M4 = 108 – 24x4 - (12-x4)3/12
ME = 108 – 24(6) - (12-6)3/12 = -54.0 k-ft
M5 = -6.0 x5 - x5
3 / 12 => MF = -6.0 (0) – (0)3 / 12 = 0
18

SFD AND BMD BY AREA METHOD
Consider a differential portion extracted from a beam as shown in the figure.V and M are shear
force and bending moment at the left end of the element and V+dV and M+dM are the shear
force and bending moment at the right end of the element.
Now apply equilibrium conditions to the free body diagram of the element.
ΣFY = 0 => V – (V+dV) +wdx = 0 => dV = wdx - - - - (a)
=> dV/dx = w - - - - - (b)
ΣMLeft face = 0 => -M + (M + dM) – (V+dV) dx + (wdx) dx/2 = 0
=> dM –V dx – dV dx – 0 = 0 => dM = Vdx - - - - (c)
=> dM/dx = V - - - - - - (d)
19

SFD AND BMD BY AREA METHOD (CONT..)
1. From equation (a), we can say that change in shear forces between two sections is equal to
the load area between those two sections.
2. From equation (b), we can say that the slope of shear force diagram at a certain section is
equal to the load intensity at that section.
3. From equation (c), we can say that change in bending moments between two sections is
equal to the shear area between those two sections.
4. From equation (d), we can say that the slope of bending moment diagram at a certain section
is equal to the magnitude of shear force at that section.
5. From left side, upward loads (concentrated or variable) are taken as positive and downward
loads are taken as negative.
6. From left side, clockwise moments are taken as positive.
7. As dM/dx = V, bending moment will be maximum where shear force is zero.
20

AREA CALCULATIONS
For first degree slope (straight line):
A1 = (b x h)/2
A2 = (b x h)/2
For second degree slope:
A1 = (b x h)/3
A2 = 2(b x h)/3
For n degree slope:
A1 = (b x h)/(n+1)
A2 = n(b x h)/(n+1)
Note: In all of the above formula, the slope at end or start must be equal
to zero
21

SLOPE OF CURVES
Following are different kind of slopes:
1. Zero slope
2. Constant Positive Slope
3. Constant Negative Slope
4. Positively increasing
5. Positively decreasing
6. Negatively increasing
7. Negatively decreasing
22

Example 3: Draw shear force and bending diagrams of the given beam using area method.
Solution:
ΣMA = 0 => VD x 30 +100 – (2 x 15) (22.5) – 10 (40) = 0 => VD = 32.5 k
ΣFY = 0 => VA + 32.5 + –(2 x 15) - 10 = 0 => VA = 7.5 k
23

Example 3 (Cont..):
Step No.2: Draw shear force diagram
VA = +7.5 k
(ΔV)AB = Load Area = 0
VB = VA + (ΔV)AB = +7.5 + 0 = +7.5 k
(dV/dx)AB = w = 0, Hence the slope is zero.
VC = VB + (ΔV)BC = +7.5 + 0 = +7.5 k
(dV/dx)BC = w = 0, Hence the slope is zero.
VD, left = VC + (ΔV)CD = +7.5 + (-2.0 x 15) = -22.5 k
(dV/dx)CD = w = -2.0, Hence the slope is negative and constant (straight line).
VD, Right = VD, Left + 32.5 = -22.5 + 32.5 = +10.0 k
VE = VD + (ΔV)DE = +10.0 + 0 = +10.0 k
(dV/dx)DE = w = 0 , Hence the slope is zero.
for UDL the point of zero shear is calculated as: Δx = ΔV/w = 7.5/2 = 3.75 ft.
24

Example 3 (Cont..):
Step No.3: Draw bending moment diagram
MA = 0
(ΔM)AB = Shear Area = +7.5 x 7.5 = 56.25 k-ft
MB, Left = MA + (ΔM)AB = 0 + 56.25 = 56.25 k-ft.
(dM/dx)AB = V = +7.5, Hence the slope is positive and constant.
There is concentrated moment at B, therefore:
MB, Right = MB, Left + (-100) = 56.25 -100 = -43.75 k-ft
MC = MB, Right + (ΔM)BC = -43.75 + 7.5 x 7.5 = +12.5 k-ft
(dM/dx)BC = V = +7.5, Hence the slope is positive and constant.
25

Example 3 (Cont..):
Step No.3: Draw bending moment diagram
Bending moment will be maximum where shear force is zero, i.e
3.75 ft right of point C
Mmax = MC + (ΔM)CC’ = +12.5 + (7.5 x 3.75/2) = 26.5625 k-ft
(dM/dx)CC’ = +Variable, Hence the slope is positive and variable.
MD = Mmax + (ΔM)C’D = +26.5625 + (-22.5 x 11.25/2) = -100 k-ft
(dM/dx)C’D = -Variable, Hence the slope is negative and variable.
VE = VD + (ΔV)DE = -100 + (10 x 10) = 0
(dV/dx)DE = V = +10 , Hence the slope is positive and constant.
26

Example 4: Analyze the given determinate beam and draw shear force and bending diagrams
using area method.
Solution:
ΣMB, Left = 0 => VA x 10 + 60 – (2 x 10) (10/2) = 0 => VA = 4.0 k
ΣMF = 0 => 4.0 x 32 + 60 - (2 x 10) (27) -5.0 (16) + VD x 12 – (3 x 12 / 2) (6) = 0 => VD = 45 k
ΣFY = 0 => 4.0 + 45 + VF –(2 x 10) – 5.0 – (3 x 12 / 2) = 0 => VF = -6 k
27

Example 4 (Cont..):
Step No.2: Draw shear force diagram.
VA = +4.0 k
(ΔV)AB = Load Area = -2 x 10 = -20 k
VB = VA + (ΔV)AB = 4.0 + (-20) = -16.0 k
As the load is negative and constant (-2.0 k/ft), the slope
of shear force diagram will be negative and constant (-2.0
k/ft).
(ΔV)BC = Load Area = 0
VC, Left = VB + (ΔV)BC = -16.0 + 0 = -16.0 k
As the load is zero, the slope of shear force diagram will
be zero.
28

Example 4 (Cont..):
Step No.2: Draw shear force diagram (Cont..).
The point of zero shear is calculated from (ΔV = w dx):
dx = ΔV/w = 4.0/2.0 = 2.0 ft
VC, Right = VC, Left + (-5) = -16 + (-5) = -21.0 k
(ΔV)CD = Load Area = 0
VD, Left = VC, Right + (ΔV)CD = -21 + 0 = -21.0 k
As the load is zero, the slope of shear force diagram will
be zero
29

Example 4 (Cont..):
Step No.2: Draw shear force diagram (Cont..).
VD, Right = VD, Left + (45) = -21 + (45) = +24.0 k
(ΔV)DE = Load Area = (-3 x 6)/2 = -9.0 k
VE = VD, Right + (ΔV)DE = 24.0 + (-9.0) = +15.0 k
As the load is negatively increasing from 0 to -3 k/ft, the
slope of shear force diagram will be negatively increasing
from 0 to -3 k/ft.
(ΔV)EF = Load Area = (-3 x 6)/2 = -9.0 k
VF = VE + (ΔV)EF = +15.0 + (-9.0) = +6.0 k
As the load is negatively increasing from -3 to 0 k/ft, the
slope of shear force diagram will be negatively decreasing
from -3 to 0 k/ft.
30

Example 4 (Cont..):
Step No. 3: Draw bending moment diagram.
MA = +60.0 k-ft
(ΔM)AA’ = Shear Area = (+4 x 2)/2 = +4.0 k-ft
MA’ = MA + (ΔM)AA’ = +60.0 + (+4.0) = +64.0 k-ft
As shear force is +ve and decreasing, the slope of bending moment
diagram will be positively decreasing.The slope starts at 4.0 and
ends at zero.
(ΔM)A’B = Shear Area = (-16 x 8)/2 = +64.0 k-ft
MB = MA’ + (ΔM) A’B = +64.0 + (-64.0) = 0
As shear force is negatively increasing, the slope of bending
moment diagram will also negatively increasing.The slope starts at
zero and ends at -21.
31

Example 4 (Cont..):
Step No. 3: Draw bending moment diagram (Cont..).
(ΔM)BC = Shear Area = -16 x 6 = -96.0 k-ft
MC = MB + (ΔM) BC = 0 + (-96.0) = -96.0 k-ft
As shear force is constant (-16 k), the slope of bending
moment diagram will be negative constant (-16 k).
(ΔM)CD = Shear Area = -21 x 4 = -84.0 k-ft
MD = MC + (ΔM)CD = -96 + (-84.0) = -180.0 k-ft
As shear force is constant (-21 k), the slope of bending
moment diagram will be negative constant (-21 k).
32

Example 4 (Cont..):
Step No. 3: Draw bending moment diagram (Cont..).
(ΔM)DE = Shear Area = 15 x 6 + 2 (9 x 6)/3 = +126.0 k-ft
ME = MD + (ΔM)DE = -180.0 + (126.0) = -54.0 k-ft
As shear force is positively decreasing with 2nd degree, the
slope of bending moment diagram will be positively
decreasing.The slope will starts at +24.0 k and ends at +15.0
k.
(ΔM)ED = Shear Area = 6 x 6 + (9 x 6)/3 = +54.0 k-ft
MF = ME + (ΔM) EF = -54.0 + (+54.0) = 0
As shear force is positively decreasing with 2nd degree, the
slope of bending moment diagram will be positively
decreasing.The slope will starts at +15.0 k and ends at +6.0
k.
33

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