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Determine, using a table to lay out your calculations, the position of the centroid for the cross sectional area. Quote x bar and y bar in mm from the bottom left hand corner of the area.

Method of Joints and Sections

a1 = 3610 x 722 = 2,606,420 mm2; a2 = 5776 x 722 = 4,170,272 mm2; a3 = 2527 x 361 = 912,247 mm2

a1 + a2 + a3 = 2606420 + 4170272 + 912247 = 7,688,939 mm2

x1 = ; x2 = ; x3 =  

y1 = ; y2 = ; y3 =

a1x1

4,704,588,100 mm3

a1y1

16,936,517,160 mm3

a2x2

1,505,468,192 mm3

a2y2

13,549,213,728 mm3

a3x3

1,152,624,084.5 mm3

a3y3

164,660,583.5 mm3

Total

 7362680376.5 mm3

30650391471.5 mm3

Thus  and

The  and  are from the bottom left hand corner of the area/drawing.

First, the beam is separated into two sections at point P (where the internal hinge is located). The two beam sections are then solved separately (independently). It is also important to note that an internal pin (hinge) provides vertical reaction.

8 x UC = 8 x 361 = 2888 KN/m

15 x UC = 15 x 361 = 5415 KN

First section of the beam (the beam on the right hand side):

Summation of moments at P, ;

-5VC + 3(5415 sin 45) = 0; 5VC = 3(5415 sin 45); VC = 2297.39 KN

Summation of moments at C, ;

5VP – 2(5415 sin 45) = 0; 5VP = 2(5415 sin 45); VP = 1531.59 KN

Second section of the beam (the beam on the left hand side):

Summation of moments at P, ;

6VA – (2888 x 4 x 4) + (2VB) = 0

6VA – 46208 + 2VB = 0; 6VA + 2VB = 46208

Summation of moments at A, ;

(2888 x 4 x 2) – 4VB + 6VP = 0

23104 – 4VB + 6(1531.59) = 0; 23104 – 4VB + 9189.54 = 0

4VB = 32293.54; VB = 8073.385 KN

Summation of forces in y-direction (vertical forces), ;

VA – (2888 x 4) + VB – VP = 0; VA – 11552 + 8073.385 – 1531.59 = 0

VA – 5010.205; VA = 5010.205 KN

The summary of reactions at A, B and C are provided in the table below

Reaction

Value

A

5010.205 KN

B

8073.385 KN

C

2297.39 KN

Calculation of reactions at the supports of frame

Summation of forces in x-direction: ;

HA + 361 KN = 0; HA = -361 KN

HA = 361 KN ← (acting towards the left hand side)

Summation of forces in y-direction: ;

VA – 1083 – 722 + VB = 0; VA + VB = 1805 KN

Summation of moments at A: ;

(1083 x 1.5) + (722 x 4.5) + (361 x 3) – (6VB) = 0

1624.5 + 3249 + 1083 – 6VB = 0; 6VB = 5956.5; VB = 992.75 KN ↑ (acting upwards).

But VA + VB = 1085 KN; VA = 1085 – VB; VA = 1085 – 992.75 = 812.25 KN ↑ (acting upwards)

Calculation of Member Forces Using Method of Joints

The summary of reaction forces is provided in the table below

Reaction

Value

HA

361 KN ←

VA

812.25 KN ↑

VB

992.75 KN ↑

This methods involves calculating member forces at each joint separately (SkyCiv Engineering, 2015). This is because it is assumed that the joints connect members in a way that rotation is allowed hence the members can only carry either compression or tension axial force (Krenk & Hogsberg, 2013).

Joint A:

Summation of forces in y-direction: ;

VA + AC sin 63.43 = 0; 812.25 + AC sin 63.43

AC sin 63.43 = -812.25; ; AC = 908.16 KN (compression)

Summation of forces in x-direction: ;

-HA + AF + AC cos 63.43 = 0

-361 KN + AF + (-908.16) cos 63.43 = 0; AF = 361 + 406.21 = 767.21 KN (tension)

Joint C:

Summation of forces in y-direction: ;

-1083 – AC sin 63.43 – CF sin 63.43 = 0

-1083 – (-908.16 sin 63.43) – CF sin 63.43 = 0

-1083 + 812.25 – CF sin 63.43 = 0; CF sin 63.43 = -270.75;

; CF = 302.72 KN (compression)

Summation of forces in x-direction: ;

CD + CF cos 63.43 – AC cos 63.43 = 0

CD + (-302.72) cos 63.43 – (-908.16) cos 63.43 = 0

CD – 135.40 + 406.21 = 0; CD = 135.40 – 406.21 = -270.81 KN; CD = 270.81 KN (compression)

Joint F:

Summation of forces in y-direction: ;

CF sin 63.43 + FD sin 63.43 = 0

CF sin 63.43 = -FD sin 63.43

CF = - FD; FD = -CF = - (-302.72)

FD = 302.72 KN (tension)

Summation of forces in x-direction: ;

FB + FD cos 63.43 – CF cos 63.43 – AF = 0

FB + 302.72 cos 63.43 – (-302.72) cos 63.43 – 767.21 = 0

FB + 135.40 + 135.40 – 767.21 = 0

FB = 767.21 – 135.40 – 135.40 = 496.41 KN (tension)

The summary of member forces is provided in the table below

Member

Force

Sense

AC

908.16 KN

Compression

AF

767.21 KN

Tension

CD

270.81

Compression

CF

302.72 KN

Compression

FD

302.72 KN

Tension

FB

496.41 KN

Tension

Forces in members using method of sections

It starts with cutting the truss or frame into 2 separate sections, ensuring that the cut runs through all the members that are being targeted. This is followed by drawing free body diagram for one of the sections and using equations of equilibrium to determine member forces (Moore, (n.d.)).

The section chosen cuts through members DE, DB and FB

It is assumed that the right hand of the section is in equilibrium

Summation of moments at B to the right hand side, ;

Calculation of Member Forces Using Method of Sections

(3 x 361) – (3 x DE) = 0; 1083 – 3DE = 0; DE = 361 KN (tension)

Summation of moments at D to the right hand side, ;

(3 x FB) – (1.5 x VB) = 0; 3FB – (1.5 x 992.75) = 0

3FB = 1489.125; FB = 496.375 KN (tension)

Summation of moments at E to the left hand side, : ;

(1.5 x VB) + (3 x FB) + (3 x DB cos 63.43) + (1.5 x DB sin 63.43) = 0

(1.5 x 992.75) + (3 x 496.375) + 1.34187DB + 1.34158DB = 0

1489.125 + 1489.125 + 2.68345DB = 0

2.68345DB = -2978.25; DB = -1109.8586 KN; DB = 1110 KN (compression)

The summary of member forces is provided in the table below

Member

Force

Sense

DE

361 KN

Tension

DB

1110 KN

Compression

FB

496.4 KN

Tension

Calculating second moment of area

The sketch of cross is as follows:

a1 = 200 x 15 = 3,000 mm2; a2 = 356 x 171 = 60,876 mm2; a3 = 200 x 15 = 3,000 mm2

a1 + a2 + a3 = 3,000 + 60,876 + 3,000 = 66,876 mm2

x1 = ; x2 = ; x3 =  

a1x1

300,000 mm3

a2x2

6,087,600 mm3

a3x3

300,000 mm3

Total

 6,687,600 mm3

Thus

h1 = 100 – 7.5 = 92.5 mm

h2 = 193 – 100 = 93 mm

h3 = 378.5 mm

Ixx =

Ixx =

Ixx = 56250 + 25668750 + 642931728 + 526516524 + 56250 + 429786750

1,625,016,252 mm4 

Calculating maximum UDL

Area = a1 + a2 + a3 = 3,000 + 60,876 + 3,000 = 66,876 mm2

Force = Stress x Area = 165 N/mm2 x 66876 mm2 = 11,034.54 KN

UDL span = 6 m = 6,000 mm

Why 200 x 15 mm plates are not required over the whole span

This is because the web of the I section beam is much wider and longer hence the load can be supported directly on the top center of the beam. As a result, the plates are not needed through the span of the beam.

First, the beam is separated into two sections at point P (where the internal hinge is located). The two beam sections are then solved separately (independently). It is also important to note that an internal pin (hinge) provides vertical reaction.

First section of the beam (the one on the right hand side):

Summation of moments at P, ;

-6VC + 4 (60 sin 45) = 0; 6VC = 4(60 sin 45); VC = 28.28 KN

Summation of moments at C, ;

-6VP – 2(60 sin 45) = 0; -6VP = 2(60 sin 45); VP = -14.14 KN; VP = 14.14 KN ↑

Calculating Second Moment of Area

Second section of the beam (the beam on the left hand side):

Summation of moments at P, ;

8VA – (4.5 x 75) + (2VB) = 0

8VA – 337.5 + 2VB = 0; 8VA + 2VB = 337.5

Summation of moments at A, ;

(3.5 x 75) – (6 x VB) + 8VP = 0

262.5 – 6VB + 8(14.14) = 0; 262.5 – 6VB + 113.12 = 0

6VB = 375.62; VB = 62.60 KN

Summation of forces in y-direction (vertical forces), ;

VA – 75 + VB – VP = 0; VA – 75 + 62.60 – 14.14 = 0

VA – 26.53; VA = 26.53 KN

The summary of reactions at A, B and C are provided in the table below

Reaction

Value

A

26.53 KN

B

62.60 KN

C

28.28 KN

Point of contraflexure is a point on the bending moment (BMD) of a beam where the BMD intersects with or cuts the x-axis i.e. where the BMD changes sign from +ve to –ve or from –ve to +ve (Sharma, (n.d.)). Therefore point of contraflexure is determined from BMD. This involves calculating bending moments along the beam and identifying the point where the bending moment changes sign.

VA = 51 KN; VB = 78 KNm

MA = 0 KNm

M1 = (51 x 1) – (15 x 1 x 0.5) = 51 – 7.5 = 43.5 KNm

M2 = (51 x 2) – (15 x 2 x 1) = 102 – 30 = 72 KNm

M3 = (51 x 3) – (15 x 3 x 1.5) = 153 – 67.5 = 85.5 KNm

M4 = (51 x 4) – (15 x 4 x 2) = 204 – 120 = 84 KNm

M5 = (51 x 5) – (15 x 5 x 2.5) – (25 x 1) = 255 – 187.5 – 25 = 42.5 KNm

M6(B) = (51 x 6) – (15 x 6 x 3) – (25 x 2) = 306 – 270 – 50 = -14 KNm

M7 = 7 x 1 x 0.5 = 3.5 KNm

M8 = 0 KNm

The sketch of BDM of the beam is as follows

From the above calculations, the sign of bending moments changes between the point when the point load is acting and the support at B. Assuming that the point where the bending moment changes sign (point of contraflexure) is P, which is at a distance x from support B, then the value of x can be determined as follows:

Summation of moments at point M to the right hand side:

(15 * x * 0.5x) – (VB * x) + (7 * 2 * (x + 1)) = 0

7.5x2 – 78x + 14(x + 1) = 0

7.5x2 – 78x + 14x + 14 = 0

7.5x2 – 64x + 14 = 0

x = 8.31 m or x = 0.225 m

Since x cannot be greater than 2m, it means that x = 0.225 m

Therefore the point of contraflexure from the left hand support A is: 4m + (2m – 0.225m)

= 4m + 1.775m = 5.775 m

Summation of forces in x-direction: ;

HA + 20 KN = 0; HA = -20 KN

HA = 20 KN ← (acting towards the left hand side)

Summation of forces in y-direction: ;

VA – 30 – 40 + VB = 0; VA + VB = 70 KN

Summation of moments at A: ;

(20 x 3) + (30 x 4.5) + (40 x 7.5) – (6VB) = 0

60 + 135 + 300 – 6VB = 0; 6VB = 495; VB = 82.5 KN ↑ (acting upwards).

But VA + VB = 70 KN; VA = 70 – VB; VA = 70 – 82.5 = -12.5 KN

VA = 12.5 KN ↓ (acting downwards)

The summary of reaction forces is provided in the table below

Reaction

Value

HA

20 KN ←

VA

12.5 KN ↓

VB

82.5 KN ↑

Calculation of member forces by method of joints

Joint A:

Summation of forces in y-direction: ;

VA + AC sin 63.43 = 0; 75 + AC sin 63.43

AC sin 63.43 = -75; ; AC = 83.86 KN (compression)

Summation of forces in x-direction: ;

HA + AF + AC cos 63.43 = 0

0 + AF + (-83.86) cos 63.43 = 0; AF = 37.51 KN (tension)

Joint C:

Summation of forces in y-direction: ;

-75 – AC sin 63.43 – CF sin 63.43 = 0

-75 – (-83.86 sin 63.43) – CF sin 63.43 = 0

-75 + 75 – CF sin 63.43 = 0; CF sin 63.43 = 0;

; CF = 0 KN

Summation of forces in x-direction: ;

CD + CF cos 63.43 – AC cos 63.43 = 0

CD + (0) cos 63.43 – (-83.86) cos 63.43 = 0

CD – 0 + 37.51 = 0; CD = -37.51 KN; CD = 37.51 KN (compression)

Joint F:

Summation of forces in y-direction: ;

CF sin 63.43 + FD sin 63.43 = 0

CF sin 63.43 = -FD sin 63.43

CF = - FD; FD = -CF = - (0)

FD = 0 KN

Summation of forces in x-direction: ;

FB + FD cos 63.43 – CF cos 63.43 – AF = 0

FB + 0(cos 63.43) – 0 (cos 63.43) – 37.51 = 0

FB + 0 – 0 – 37.51 = 0

FB = 37.51 KN (tension)

The summary of member forces in two decimal places is provided in the table below

Member

Force

Sense

AC

83.9 KN

Compression

AF

37.5 KN

Tension

CD

 37.5 KN

Compression

CF

0 KN

-

FD

0 KN

-

FB

37.5 KN

Tension

Forces in members using method of sections

The section chosen cuts through members CD, FD and FB

It is assumed that the right hand of the section is in equilibrium

Summation of moments at F to the right hand side, ;

Assuming that force CD is in tension

(1.5 x 75) – (3 x VB) – (3 x CD) = 0; 112.5 – 3(75) – 3CD = 0; 112.5 – 225 – 3CD = 0

-112.5 = 3CD; CD = -37.5 KN

Since CD was assumed to be in tension but the value obtained is -37.5 KN, it means that it is a compression force.

Therefore CD = 37.5 KN (compression)

Summation of moments at D to the right hand side;;

Assuming that force FB is in tension

-(1.5 x VB) + (3 x FB) = 0; -(1.5 x 75) + 3FB = 0; -112.5 + 3FB = 0;

3FB = 112.5; FB = 37.5 KN (tension)

Summation of moments at B to the left hand side, ;

-(1.5 x 75) – (3 x CD) – (3 x FD cos 63.43) – (1.5 x FD sin 63.43) = 0;

-112.5 – 3CD – 1.3419FD – 1.3419FD = 0

-112.5 – (3 x -37.5) – 2.6838FD = 0

-112.5 + 112.5 – 2.6838FD = 0

2.6838FD = 0; FD = 0 KN

The summary of member forces is provided in the table below

Member

Force

Sense

CD

37.5 KN

Compression

FD

0 KN

-

FB

37.5 KN

Tension

The values of CD, FD and FB above are the same as those obtained in section 6(b) using method of joints.

Question 7: Shear force diagram (SFD) and bending moment diagram (BMD)

Summation of moments at A, ;

(10 x 4 x 2) + (25 x 7) – (9VB) + (15 x 3.5 x 10.75) = 0

80 + 175 – 9VB + 564.375 = 0

819.375 = 9VB; VB = 91 KN ↑

Summation of forces in y-direction (vertical forces), ;

VA – (10 x 4) – 25 + VB – (15 x 3.5) = 0

VA – 40 – 25 + 91 – 52.5 = 0

VA = 26.5 KN

The SFD is as shown below

The BMD is as shown below

MA = 0 KNm

M1 = (26.5 x 1) – (10 x 1 x 0.5) = 26.5 – 5 = 21.5 KNm

M2 = (26.5 x 2) – (10 x 2 x 1) = 53 – 20 = 33 KNm

M3 = (26.5 x 3) – (10 x 3 x 1.5) = 79.5 – 45 = 34.5 KNm

M4 = (26.5 x 4) – (10 x 4 x 2) = 106 – 80 = 26 KNm

M5 = (26.5 x 5) – (10 x 4 x 3) = 132.5 – 120 = 12.5 KNm

M6 = (26.5 x 6) – (10 x 4 x 4) = 159 – 160 = -1 KNm

M7 = (26.5 x 7) – (10 x 4 x 5) = 185.5 – 200 = -14.5 KNm

M8 = (26.5 x 8) – (10 x 4 x 6) – (25 x 1) = 212 – 240 – 25 = -53 KNm

MB = 15 x 3.5 x 1.75 = -91.875 KNm

M10 = 15 x 2.5 x 1.25 = -46.875 KNm

M11 = 15 x 1.5 x 0.75 = -16.875 KNm

References

Krenk, S. & Hogsberg, J., 2013. Truss Structures. In: Statics and Mechanics of Structures. Dordrecht: Springer, pp. 39-89.

Moore, J., (n.d.). The Method of Sections. [Online]
Available at: https://adaptivemap.ma.psu.edu/websites/structures/method_of_sections/methodofsections.html
[Accessed 28 July 2018].

Sharma, S., (n.d.). Procedure for drawing shear force and bending moment diagram:. [Online]
Available at: https://nptel.ac.in/courses/112107146/lects%20&%20picts/image/lect23%20and%2024/lecture%2023%20and%2024.htm
[Accessed 28 July 2018].

SkyCiv Engineering, 2015. Tutoral for truss method of joints. [Online]
Available at: https://skyciv.com/tutorials/tutorial-for-truss-method-of-joints/
[Accessed 28 July 2018].

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