300 kN

200

150

A= 13744.468

21.8

mm2

s= N/mm2

50 mm

25 mm

100 kN

A= 490.874

203.7

mm2

s= N/mm2

A= 1963.5

50.9

mm2

s= N/mm2

Maximum Stress = 200

25 mm

15 mm 10 mm

12mm Dia.

Cross Sectional Area of Link = 150

F max = 30000 N30 kN

Cross Sectional Area of Eye = 130

F max = 26000 N26 kN

N/mm2

mm2

mm2

F kN

1Tie

L= 5 m

N = 75

A= 500E= 200

1507.500E-04

Elongation= 3.75 mm

2Strut

L= 1.5 m

N = -30

A= 250E= 70

-120-1.71E-03

Elongation= -2.57 mm

mm2

kN/mm2

s = N/mm2

e=

mm2

kN/mm2

s = N/mm2

e=

kN Tension

kN Compression

Stress =

Determine the Percentage change in the volume of the cube

Material E= 1.00E+05Poisson's ratio = 0.25

Assuming cube size is 1000x1000x1000 (could have used 1x1x1)

Delta Long. Delta Lat.

-8.00E-04 -8.00E-01 mm 2.00E-04 2.00E-01

N/mm2

Long. e Lat. e

-80

Original Vol. New Vol. Delta Vol.

mm 1000000000 999599719.968 0.04%

N/mm2

mm3 mm3

1000 mm 50 mm Diameter

Calculate :1. The vertical movement of B2. The total extension3.The Change in lateral dimensions of AB and BC4. The change in volume of the hanger

1000 mm 35 mm Diameter

135 KN 135000 N

Material Steel E= 200Poisson's ratio = 0.3

Stress Delta Long. Delta Lat. Original Vol.

A-B 68.7549 0.000344 0.34377 mm -0.0001031 -0.0051566 mmB-C 140.316 0.000702 0.70158 mm -0.0002105 -0.0073666 mm

Total 1.0454 mm

KN/mm2

Long. e Lat. eN/mm2

3.The Change in lateral dimensions of AB and BC

Original Vol. New Vol.

1963495 1963765962112.75 962382.51

2925608 2926148 -539.64

mm3 mm3

mm3

Stress in Timber Beam (4x2)due overweight Lecturer at midspam

Breadth 50

100 Depth

Load 1962 N

I = 4E+06where b is breadth and d is depth

M 1962 Nm

z max = 50 mm

mm4 I=bd3/12

y y

z max

z max

23.54

23.54 Compression

Neutral Axis

23.54 Tension

s max = N/mm2

N/mm2

N/mm2

Stress in Timber Beam (4x2)due overweight Lecturer at midspam

Me at Midspan 200 kg

Span = 4 m

b is breadth and d is depth

Stress in Concrete Beam due to self weight

Breadth 300

800 Depth

Density Concrete 2400

w 5650.56 N/m5.65056 kN/m

M= 70.632 kNm 70632000 Nmm

I = 1.280E+10where

z max = 400 mm

W top/bot= 32000000

kg/m3

mm4 I=bd3/12

mm3

y y

z

z max

2.21

2.21 Compression

Neutral Axis

2.21 Tension

s max = N/mm2

N/mm2

N/mm2

Span 10 m

b is breadth and d is depth

T Beam subject t 7.00 kNm Moment

100

28.75 15

28.7578.750

100

15

A1 1500 h1 28.75A2 1500 h2 28.75A 3000

Moments of Areas about the Base3000 y' = 1500 50

y' = 78.75 mm

Iyy1 = 2.813E+04Iyy2 = 1.250E+06

Iy'y' = 3.758E+06

1.037E+05

4.772E+04

67.53

146.69

mm4

Wtop = mm3

Wbot = mm3

s top = N/mm2

s bot = N/mm2

y’ y’

67.53

Neutral

146.69

= 7.00L (say) = 10 mw = 0.56

mmmm

+ 1500 107.5

wl2/8

y’

Compression

Axis

Tension

N/mm2

N/mm2

span continuous over one support plus cantilever made from steel box section given

175 kN

2 m 7 m 3

RA + RC = 225 kN

M about A

175 x 2 + 50 x

RC = 105.56 kNRA = 119.44 kN

Moment about end of Cantilever = 0

Check = 0

Therefore

M at B 238.89 kNm

M at C -150 kNm

BMD -150-

B

CA

+

238.89 kNm

Section

250 mm 300

I = 3E+08

zmax top/bot = 150 mm

s max = 102.53

-102.5 N/mm2 Compression

102.5 N/mm2 Tension

mm4

N/mm2

Neutral Axis

span continuous over one support plus cantilever made from steel box section givenSection

50 kN

250 300mm mm

m

12 = RC x 9

kNm

C

mm

Given that the maximum permissible stress 165What is the maximum value of F

70 mm

7

64 mm54.5 mm

102

Standard I section Iyy = 21854.5 mm

7

Underneath the point load

Use paraleel Axis Theorem to Calculate I value at centre

Part Icc Area h

Top Plate 2.00E+03 490 54.5 1455423Standard Beam 2.18E+06 .... 0 0Bottom Plate 2.00E+03 490 54.5 1455423

2.18E+06 2.91E+06 Total I =

zmax = 58 mm

W = 8.78E+04

M = 1.45E+07 Nmm

N/mm2

cm4

Ah2

mm3

14.49 kNm

F = 9.66 kN

Checking Curtailment of Flange Plate

At this point

I = 2.18E+06

zmax = 51 mm

W = 4.27E+04

Based on F = 9.66 M @ D = 9.66264 kNm

Hence the stress at this point = 226.05 which would cause failure

permissible moment 7.05E+06 Nmm7.05 kNm

Based on M = 7.0529 F = 7.05 kN

mm4

mm3

N/mm2

F2 m

A C B

mm

D E

6 m

mm

mm

5.1E+06 mm4

which would cause failure

100 x 50 x 10 PFC (Parallel Flange Channel)s with 10mm cover plates Neglecting self-weight calculate the maximum distributed load on a 10m span ….. Given :

200 N/mm2

10 mm

10 mm

100 mm

10 mm

Part Icc Area hTop Plate 9.17E+03 1100 55 3327500Standard Sections 4.16E+06 .... 0 0Bottom Plate 9.17E+03 1100 55 3327500

4.18E+06 6.66E+06

Total I = 1.1E+07

M = 3.61E+07 Nmm

= 36.1 kNm

smax =

Ah2

mm4

y’y’

=

w = 2.9 kN/m

wl2/8

PFC (Parallel Flange Channel)s with 10mm cover plates Neglecting self-weight calculate the maximum distributed load on a 10m span ….. Given :

I of 1 Channel 208 cm4

w kN/m

10 m Span

y’

Neglecting self-weight calculate the maximum distributed load on a 10m span ….. Given :

Given that:Channel 38 264Universal Beam (UB) 60.8 9500

total 9880 97640000

What are the maximum tensile and compressive stresses on the combined/compound section

8 mm 26.5To Centroid of Channel

Y

310 mm

155 mm

Determine position of Centroid of Compound section

Taking Moments of areas about the Base A z' = Sum(Ai z'i)

9880 z' = 3800 xz' = 207.5 mmztop = 110.5 mm

zbot = 207.5 mm

Use Parallel Axis Theorem to Calculate I value of compound at new centriod

Part Icc Area h

Channel 2.64E+06 3800 84 26812800Universal Beam 9.50E+07 6080 52.5 16758000

Total 9.76E+07 4.36E+07

Area (cm2) Icc (cm4)

mm2

Ah2

Total I = 1.412E+08

70.43 Compression

132.25 Tension

s top = N/mm2

s bot = N/mm2

What are the maximum tensile and compressive stresses on the combined/compound section 20 kN/m

A

mmTo Centroid of Channel

6 m

B.M.D

Mmax = 90 kNm

Mmax = 90000000 Nmm

291.5 + 6080 x 155

Use Parallel Axis Theorem to Calculate I value of compound at new centriod

mm4

mm4

B

Given that:Span = 8 m With Superimposed Load as IndicatedThe density of the Fibre Reinforced Concrete =

What are the maximum tensile and compressive stresses on the section due to a combiantion of self weight and imposed Loa

300 mm

A 60

Fibre Reincorced Concrete I-Beam300 460 mm B

40 mm

Y

z'

C

500 mm

Determine position of Centroid of Compound sectionTaking Moments of areas about the Base

A z' = Sum(Ai z'i)

A 18000 430 7740000B 12000 250 3000000C 50000 50 2500000

Total 80000 13240000

A z' = Sum(Ai z'i)

80000 z' = 13240000z' = 165.50 mm

mm2

Use Parallel Axis Theorem to Calculate I value of A,B & C about this centriod

Part Icc Area hA 5.4000E+06 18000 264.50 1.26E+09B 9.0000E+07 12000 84.50 8.57E+07C 4.1667E+07 50000 115.50 6.67E+08

Total 1.3707E+08 2.0120E+09

Total I =

Bending Moment due to self weight

Volume of Beam = Area x Length = 80,000 x

Therefore Weight = 0.08 x 2400 x 9.81

= 1.88352 x

= 15.07 Knm

Bending Moment due to self weight

Reaction VA = VB = 6.25 x 4 / 2

Therefore Maximum Moment (at midspan) = 12.5 x 4

= 37.50 Knm

Stresses due to combined Moment

Combined Maximum Moment (at midspan) = 15.07 + 37.50

Due to Combined loading the maximum tensile and compressive stresses on the section

7.20 Compression

4.05 Tension

Ah2

Therefore Maximum moment at midspan = wl2/8

s top = N/mm2

s bot = N/mm2

With Superimposed Load as Indicated2400

What are the maximum tensile and compressive stresses on the section due to a combiantion of self weight and imposed Loa 6.25 kN/m

A

mm

2 m 4 m

8 m

Fibre Reincorced Concrete I-Beam

Y

B.M.D SDL

100 mm Mmax (SDL)= 37.5 kNm

Mmax (SW)= 15.06816 kNm

kg/m3

mm3

Use Parallel Axis Theorem to Calculate I value of A,B & C about this centriod

2.1490E+09

2.15E+09

1 = 0.08

= 1.88352 kN/m

8 / 8

= 12.5 kN

- 12.5 x 1

= 52.57 Knm

Due to Combined loading the maximum tensile and compressive stresses on the section

7.20E+00

4.05E+00

mm4

m3/m

2

B

2 m