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Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-1
BEAM
A horizontal or inclined structural member that is designed to resist forces acting to its axis is called a beam INTERNAL FORCES IN A BEAM Whether or not a beam will break, depend on the internal resistances built up in the beam and on the strength of material that the beam is made of. What are the internal resistences and how to find them? If the beam can still maintain in one piece without breaking under the actions of lateral loads, then the beam as a whole, and any part of it, must be in the condition of equilibrium. By means of equations of equilibrium for the plane structure, we should be able to find the internal resistance in a beam.
ΣFx = 0, ΣFy = 0 and ΣM = 0.
The internal forces for a beam section will consist of a shear force V, a bending moment M, and an axial force (normal force) N. For beams with no axial loading, the axial force N is zero. Sign Convention for Internal Forces in a Beam
Sign Convention (considering a small segment of the member): Shear Force V: Positive shear tends to rotate the segment clockwise. Moment M: Positive moment bends the segment concave upwards. (so as to ‘hold water’) Axial Force N: Tension is positive.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-2
An important feature of the above sign convention (often called the beam convention) is that it gives the same (positive or negative) results regardless of which side of the section is used in computing the internal forces.
V
N
M MN
V
V
V
N N
M M
Axial Force
Shear Force Bending Moment
POSITIVE SIGN CONVENTION
PROCEDURES FOR FINDING V, M AND N AT A BEAM SECTION 1. Identify whether the beam is a determinate or an indeterminate
structure. (This chapter focuses on the analysis of determinate beam only. Indeterminate structure requires the consideration of compatibility condition, i.e. the deformation of the structure).
2. Compute the Support Reactions
Make use of the equilibrium equations and the equations of condition if any.
3. Draw a Free-Body Diagram of the Beam Segment
Keep all external loading on the member in their exact location. Draw a free-body diagram of the beam segment to the left or right of the beam. (Although the left or right segment could equally be used, we should select the segment that requires the least computation). Indicate at the section the unknowns V, M and N. The directions of these unknowns may be assumed to be the same as their positive directions.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-3
4. Use the Equations of Equilibrium to Determine V, M & N. If the solution gives a negative value for V, M or N, this does not mean the force itself is NEGATIVE. It tells that actual force or moment acts in the reversed direction only.
5. Check the Calculations using the Opposite Beam Segment if
necessary.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-4
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-5
Example 1 Find the shear and moment on sections at 2 m and 4 m from the left end for the following beam.
Solution To analyze internal forces in the beam, it will first be necessary to calculate the reaction at one support.
ΣMD = 0, VA*7 = 40*4 + 20*2
VA = 28.571 kN
ΣFx = 0, HA = 0
ΣFy = 0, VA + VD = 20 + 40
VD = 31.429 KN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-6
A free-body diagram of the part of the beam to the left of the section at 2 m is drawn to analyze the shear and moment on the section. At x = 2m,
ΣFy = 0,
V=28.571 kN ↓
ΣM0 = 0,
M = 28.571*2 M = 57.142 kNm (Sagging) A new free-body diagram is drawn to shown V and M on the section 4 m from right of the left end. Again, both V and M are drawn in the positive direction for internal shear and moment.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-7
At x = 4m,
ΣFy = 0,
28.571- 40 = V V = -11.429 kN ↑ (Negative sign indicates that the shear force acts
upwards direction.)
ΣM0 = 0,
-28.571*4 + 40*1 + M = 0 M = 74.824 kNm (Sagging)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-8
Example 2 Determine the shear force V and the bending moment M at the section P of the overhanging beam shown.
4m
A B C DP
10 kN 15 kN4 kN/m
2m 10m 3m
HBVVB C
Solution:
No. of reactions = no. of equations of equilibrium ⇒ This is a determinate beam.
Determine reactions HB, VB and VC.
ΣX = 0, HB = 0 Take moment about B, ΣM = 0, 4*10*(10/2) + 15*13 – 10*2 – VC*10 = 0 VC = 37.5 kN ΣY = 0, VB + VC = 10 + 4*10 + 15 VB = 27.5 kN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-9
Determine V and M at P (using left free-body)
HB VB
4 kN/m10 kN
A B P HV
M
PP
P
2m 4m
∑X = 0, since HB = 0, ∴ HP = 0
∑Y = 0, VB + VP = 10 + 4*4
27.5 + VP = 26 ∴ VP = -1.5 kN
(This implies that Vp acts in downwards direction ↓)
Take moment about P, 10*6 + 4*4*4/2 + MP = VB*4 MP = 27.5*4 – 60 – 32 = 18 kNm
Determine V and M at P (using right free-body)
4 kN/m15 kN
6m 3mVC
HP
V
MP
PP
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-10
∑X = 0, HP = 0,
∑Y = 0, VC + VP = 15 + 4*6
37.5 + VP = 39 ∴ VP = +1.5 kN
(This implies that Vp acts in upwards direction ↑ as assumed)
Take moment about P, 15*9 + 4*6*6/2 + MP = VC*6 MP = 37.5*6 – 135 – 72 = 18 kNm (This implies that MP acts in the direction as indicated in the free-body diagram.)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-11
Example 3 Determine the shear force V and the bending moment M at section P of the cantilever beam.
HA
VA
MAA
B P C 3m 3m4m
40 kN
5 kN/m
Solution: Determine the support reactions
∑X = 0, HA= 0, ∑Y = 0, VA = 5*6 + 40 = 70 kN
Take moment about A, 40 * 3 + 5*6*6/2 – MA =0 MA = 210 kNm
Determine V and M at P (using left free-body)
A
B
P
4m
40 kN
5 kN/m210 kNm
70 kN
HP V
MP
P
∑X = 0, HP= 0, ∑Y = 0, VP + 70 = 40 + 5*4 VP = -10 kN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-12
Take moment about P, 40*1 + 5*4*4/2 + 210 – 70*4 - Mp = 0 MP = 10 kNm
Determine V and M at P (using right free-body)
5 kN/mHP
VP
MP
P C 2m
∑X = 0, HP= 0, ∑Y = 0, VP = 5*2 = 10 kN
Take moment about P, 5*2*1 – Mp = 0 MP = 10 kNm
*Both the left free-body and the right free-body can be used to obtain the results. However, it is noted that by using the right free-body will greatly simplified the calculations. This shows importance of choosing the appropriate free-body.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-13
SHEAR FORCE AND BENDING MOMENT DIAGRAMS By the methods discussed before, i.e. by using free-body diagrams, the magnitude and sign of the shear forces and bending moments may be obtained at many sections of a beam. When these values are plotted on a base line representing the length of a beam, the resulting diagrams are called, respectively, the shear force diagram and the bending moment diagram. Shear force and bending moment diagrams are very useful to a designer, as they allow him to see at a glance the critical sections of the beam and the forces to design for. Draftsman like precision in drawing the shear force and bending moment diagrams is usually not necessary, as long as the significant numerical values are clearly marked on the diagram. The most fundamental approach in constructing the shear force and bending moment diagrams for a beam is to use the procedure of sectioning. With some experience, it is not difficult to identify the sections at which the shear force and bending moment diagrams between these sections are readily identified after some experience and can be sketched in.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-14
Example 4 Draw the shear force and the bending moment diagrams for the beam shown.
HA VA VC
B
5m 3m
8 kN
2 kN/mA C
Solution:
∑X = 0, HA = 0 Take moment about A, 2*8*8/2 + 8*5 – VC*8 = 0 VC = 13 kN
∑Y = 0, VA + VC = 2*8 + 8 ⇒ VA + 13 = 2*8 + 8 ⇒ VA = 11 kN
2 kN/m
11 kN x
HX
V
MX
X
For 0 ≤ x ≤ 5m
∑X = 0, HX = 0 ∑Y = 0, VX + 2x = 11, ⇒ VX =11 - 2x
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-15
Take moment about the “cut”, 11x – 2x(x)/2 – Mx = 0, ⇒Mx = 11x – x2
B
5m
8 kN
2 kN/m
11 kN
x
HX
V
MX
X
For 5 ≤ x ≤ 8m
∑X = 0, HX = 0 ∑Y = 0, VX + 11 = 2x + 8, ⇒ VX = 2x - 3
Take moment about X, 11x – 2x(x)/2 – 8*(x-5) - Mx = 0, ⇒Mx = 11x – x2 –8x + 40
⇒Mx = 3x – x2 + 40
X (m) 0 1 2 3 4 5 5 6 7 8 V (kN)
11 9 7 5 3 1 -7 -9 -11 -13
M (kNM)
0 10 18 24 28 30 30 22 12 0
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-16
B
5m 3m
8 kN
2 kN/m
11 kN 13 kN
Shear Force (kN)11 9 7 5
3 1
-7-9 -11
-13
0 010
1824 28 30
12
22
Bending Moment (kNm)
A C
AB
C
CBA
+ve
+ve
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-17
Example 5 Draw the shear force and bending moment diagrams for a cantilever beam carrying a distributed load with intensity varies linearly from w per unit length at the fixed end to zero at free end.
HA
VA
MA
B
w kN/m
X
wxl
Solution: At any section distance x from the free end B,
Vx = l
wxxl
wx22
2=⎟
⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛
Mx = l
wxxl
wx632
32−=⎟
⎠⎞
⎜⎝⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-18
HA
VA
MA
B
w kN/m
X
wxl
wxl
Hx
V
Mx
xx
wl/2
V = wx /2lx 2
Shear Force Diagram
-wl /6M = -wx /6lx 3 Bending Moment Diagram
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-19
Example 6 Draw the shear force and bending moment diagrams for the beam subject to a concentrated moment M* at point C.
HB VB VA
A BC M*
La b
Solution:
∑X = 0, HB = 0
Take moment about A, VB * L = M*, ⇒ VB = M*/L (↑)
∑Y = 0, VB – VA = 0, ⇒ VA = M*/L (↓)
Take the left free-body (for 0 ≤ x < a)
A
xM /L*
HX
V
MX
X
∑X = 0, HX = 0
∑Y = 0, VX = M*/L
Take moment about the cut section, (M*/L)*(x) = MX (hogging moment)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-20
Take the left free-body (for a < x ≤ L)
A C M*
ax
V
MX
XM /L*
∑Y = 0, VX = M*/L
Take moment about the cut section, (M*/L)*(x) + MX = M* MX = M* - (M*/L)*(x) ⇒ MX = M*(L – x)/L
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-21
A BC M*
La b
M /L* M /L*
M b/L
-M a/L
AB C
-M /L*Shear Force Diagram
Bending MomentDiagram
*
*
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-22
RELATIONSHIPS BETWEEN LOAD, SHEAR FORCE AND BENDING MOMENT Consider a beam element subject to distributed load as shown below.
M
V
V+dV
M+dM
dx
q
∑Y = 0, V = q(dx) + (V+dV)
∴ qdxdV
−= ⇒ ∫ ∫−=B
A
B
AqdxdV
⇒ VB – VA = ∫−B
Aqdx
= - (area of load-intensity diagram between points A and B)
∑M = 0, -M + q(dx)(dx)/2 + (M+dM) - V(dx) = 0 Ignore the higher order terms,
we get Vdx
dM= ⇒∫ ∫=
B
A
B
AVdxdM
⇒ MB – MA = ∫B
AVdx
= area of shear force diagram between points A and B.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-23
SUMMARY OF THE RELATONSHIPS BETWEEN LOADS, SHEAR FORCE AND BENDING MOMENTS
Slope of shear force diagram at a point
=
Intensity of distributed load at that point
Change in shear between points A and B
=
Area under the distributed load diagram between points A and B.
Slope of bending moment diagram at a point.
=
Shear at that point
Change in bending moment between points A and B.
=
Area under the shear force diagram between points A and B.
Concentrated Loads Change in shear at the point of application of a concentrated load.
=
Magnitude of the load.
Couples or Concentrated Moments Change in bending moment at the point of application of a couple.
=
Magnitude of the moment of the couple.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-24
SHAPES OF SHEAR FORCE AND BENDING MOMENT DIAGRAMS
A. Beams under Point Loads only 1. Shears are constant along sections between point loads. 2. The shear force diagram consists of a series of horizontal lines. 3. The bending moment varies linearly between point loads. 4. The bending moment diagram is composed of sloped lines.
B. Beams under Uniformly Distributed Loads (UDL) only 1. A Uniformly Distributed Load produces linearly varying shear
forces. 2. The shear force diagram consists of a sloped line or a series of
sloped lines. 3. A UDL produces parabolically varying moment. 4. The bending moment diagram is composed of 2nd-order
parabolic curves.
C. Beams under General Loading 1. Section with No Load: Shear force diagram is a Horizontal Straight Line. Moment Diagram is a Sloping Straight Line. 2. At a Point Load:
There is a Jump in the Shear Force Diagram.
3. At a Point Moment: There is a Jump in the Bending Moment Diagram.
4. Section under UDL: Shear Force Diagram is a sloping straight line (1st order) Bending Moment Diagram is a Curve (2nd order parabolic)
5. Section under Linearly Varying Load Shear Force Diagram is a Curve (2nd order) Bending Moment Diagram is a Curve (3rd order)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-25
6. The Curve of the Bending Moment Diagram is 1 order above
the Curve of the Shear Force Diagram.
7. Maximum and Minimum Bending Moments occur where the Shear Force Diagram passes through the X-axis (i.e. at points of zero shear) (This characteristics is very useful in finding Max. and Min. bending moment.)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-26
Example 7 Draw the shear force and bending moment diagrams for the beam shown.
HA VA VF
AB C D E F
1.5m 3m1.5m 1.5m 1.5m
20 kN 40 kN4 kN/m
Solution: ∑X = 0, HA = 0 kN Take moment about A, 20*1.5 + 4*3*(3 + 3/2) + 40*7.5 – VF*9 = 0 ⇒ VF = 42.7 kN ∑Y = 0, VA + VF = 20 + 40 + 4*3 ⇒ VA = 29.3 kN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-27
Shear Force and Bending Moment
AB C D E F
20 kN 40 kN4 kN/m
29.3 42.7
29.3 29.3
9.3
-2.7
-42.7 -42.7
0
2.325m
Shear Force (kN)
+44.0
+14.0 +10.8
-63.9
-4.0-0.9
4458
68.8 67.9 63.9
0
Bending Moment (kNm)
A B CD
E F
A B C D E F
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-28
Example 8 Draw the shear force and bending moment diagrams for the beam shown.
HB VB VD
A B C D E
10 kN 20 kN4 kN/m
2m 2m 4m 2m Solution: ∑X = 0, HB = 0 kN Take moment about B, 20*2 + 4*8*4 – 10*2 – VD*6 = 0 ⇒ VD = 24.7 kN ∑Y = 0, VB + VD = 10 + 20 + 4*8 ⇒ VB = 37.3 kN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-29
Shear Force and Bending Moment
37.3
A B C D E
10 kN 20 kN4 kN/m
24.7
-10 -10
27.319.3
-0.7
-16.7
8
0
Shear Force (kN)
-20
+46.7
-34.7
+8
14
14
14
-20
+26.7
-80
101
Bending Moment (kNm)
A B C DE
E DC
BA
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-30
Deflected Shape of a Beam The qualitative deflected shape (also called “elastic curve”) of a beam is simply an approximate and exaggerated sketch of the deformed beam due to the given loading. The deflected shape is useful in understanding the structural behaviour. Sketching the Deflected Shape of a Beam 1. The deflected shape must be consistent with the support conditions:
(a) At a Roller Support, the vertical deflection is zero but the beam may rotate freely.
(b) At a Pin Support, the vertical and horizontal deflections are zero but the beam may rotate freely.
(c) At a Fixed Support, the vertical and horizontal deflections are zero and there is no rotation.
2. The deflected shape must be consistent with the Bending Moment
Diagram. (a) Where the moment is positive, the deflected shape is concave
upwards ( ∪ ). (b) Where the moment is negative, the deflected shape is concave
downwards ( ∩ ).
3. The transition points between positive and negative moment regions are points of zero moment. These points are called “point of inflection” or “point of contraflexure”.
4. The deflected shape must be a smooth curve except at internal hinges.
Normally, the vertical deflection at an internal hinge is not zero. 5. Quite often it is possible to sketch the deflected shape of a structure first
and then to infer the shape of the bending moment diagram from the sketch. This is useful for checking whether a bending moment diagram obtained through calculations is correct.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-31
Deflected Shape
A B C
A BC
Bending Moment
Deflected Shape
+ve moment -ve moment
point of inflection
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-32
Examples of Beam Deflected Shape
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-33
Examples of Beam Deflected Shape
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-34
Examples of Beam Deflected Shape
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-35
Examples of Beam Deflected Shape
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-36
Example 9 Construct the complete shear force and bending moment diagrams, and sketch the deflected shape for the beam shown.
30 kN/m
10 kN 20 kN
2m 2m5m
A B C D
Solution:
VHB VB C
30 kN/m
10 kN 20 kN
2m 2m5m
A B C D
∑X = 0, HB = 0 kN Take moment about B, 20*7 + 30*7*(3.5 – 2) + (30*2/2)*(5 + 2/3) - 10*2 – VC*5 = 0 ⇒ VC = 121 kN ∑Y = 0, VB + VC = 10 + 20 + 30*7 + 30*2/2 ⇒ VB = 149 kN
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-37
-10
-70
50
D
Shear Force (kN)
AB C
79
-71
20
2.63m
Bending Moment (kNm)
A B C D
2.63 m
-80-60
24
30 kN/m
10 kN 20 kN
A B C D
Deflected Shape
P.I. P.I.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-38
PRINCIPLE OF SUPERPOSITION The principle of superposition states that on a linear elastic structure, the combined effect (e.g., support reactions, internal forces and deformation) of several loads acting simultaneously is equal to the algebraic sum of the effects of each load acting individually. There are two conditions for which superposition is NOT valid. 1. When the structural material does not behave according to Hooke’s law;
that is, when the stress is not proportional to the strain. 2. When the deflections of the structure are so large that computations
cannot be based on the original geometry of the structure.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-39
Principle of Superposition
HA
VA
MA
P1P2w kN/m
L L
B.M. due to P1
B.M. due to P2
B.M. due to w
2P L
P L
2wL
1
2
2
m1
m2
m3
(a)
(b)
(c)
m1 + m2 + m3
2P L1 P L2 2wL2+ +
(d)
+
+
Complete bendingmoment diagram
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-40
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-41
SUPERPOSITION
4 kN/m 40 kNm
5 kN
15
77.5
2.5m 2.5m
4 kN/m
10
12.5
5 kN
5
25
40 kNm40
+
+
-77.5
-12.5
-52.5
-25
-40 -40
-12.5
-12.5
+
+
Superposition of loading Superposition ofbending moment (kNm)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-42
Bending Moment Diagrams by Parts
4m 2m
5 kN/m10 kN
25 kN5 kN
5 kN/m
10 kN10 kN
10 kN
15 kN5 kN
+
10 kNm
-20 kNm
-20 kNm
10 kNm -20 kNm
2.5 kNm
+
Bending Moment
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-43
Beam Deflection Beam deflection can be determined by using the following beam deflection table. The deflection of the beam is inversely proportional to the quantity EI, which is the flexural rigidity of the beam.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-44
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-45
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-46
Example 10 Use the methods of superposition to find the deflection at the free end of the following cantilever beam. EI of the beam is constant.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-47
Solution
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-48
where WT = W1 + W2
From the deflection tables, WT = PL3/3EI + wL4/8EI WT = 2(9)3/(3EI) + 0.5(9)4/(8EI) WT = 896 /(EI) kN-m3
Example 11
Use the method of superposition to find the deflection at the middle of the following simply supported beam. EI of the beam is constant.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-49
Solution
10 kN
A B
3 m
y1
A B
y2
6 m
1.5 kN/m
Total deflection at the mid-span of the beam, yM
yM = y1 + y2
From the beam deflection tables,
yM = Pa(3L2 - 4a2) / 48EI + 5wL4 / 384 EI
= 10(3)[3(12)2 - 4(3)2] / 48EI + 5(1.5)(12)4 / 384EI
= 652/EI kN-m3
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-50
Example 12
Determine the displacement at point C and the slope at the support A of the beam. EI of the beam is constant.
Solution
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-51
Total deflection at the point C of the beam, yc
yc = y1 + y2
yc = PL3 / 48EI + 5wL4 / 768EI
yc = 8(8)3 / 48EI + 5(2)(8)4 / 768EI
yc = 139/EI kN-m3
Total slope at the support A of the beam, θc
θc = θ1 + θ2
= PL2 / 16EI + 3wL3 / 128EI
= 8(8)2 / 16EI + 3(2)(8)3 / 128EI
= 56/EI (Clockwise direction)
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-52
Example 13
Use the method of superposition to find the deflection at the free end of
the following cantilever beam.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-53
Solution
The uniform loading produces a deflection at B and because the beam is
continuous and has a slope at B, an additional deflection occurs at C
equal to (L1-L2)θB as shown in the following figures.
LL 21
0.5 kN/m
CA B
-
y
y1
2
CA B
2 kN
y3
0
Total deflection at C = y1 + y2 + y3
= wL24 / 8EI + (L1 -L2 )*wL2
3 /6EI + PL13 / 3EI
= 0.5(6)4 / 8EI + 3*0.5*(6)3 /6EI + 2*(9)3 / 3EI
= 624/EI kN-m3
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-54
Example 14
A cantilever beam is subject to a uniformly distributed load of 0.5 kN/m
over the length BC as shown in the following figure. If the flexural rigidity
(EI) of the beam is 6000 kN-m2, calculate the deflection at the free end C of
the beam.
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-55
Solution
0.5 kN/m C
A
B
0.5 kN/m
CAB
0.5 kN/mC
AB
y1
y2
y3
0
Total deflection at C = y1 - y2 - y3
= 0.5(7)4 / 8EI - 0.5(2)4 / 8EI -5*0.5*(2)3 /6EI = 145.73/EI kN-m3
= 145.73/6000 m
= 0.024 m
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-56
Hong Kong Institute of Vocational Education (Tsing Yi) Higher Diploma in Civil Engineering – Structural Mechanics Chapter 4 – Analysis of Determinate Beams
Page 4-57