temperature stresses in Strength of materials

STATICALLY INDETERMINATE MEMBERS
&
THERMAL STRESSES
Chapter II 1

STATICALLY INDETERMINATE MEMBERS
Structure for which equilibrium equations are sufficient to obtain the
solution are classified as statically determinate. But for some
combination of members subjected to axial loads, the solution
cannot be obtained by merely using equilibrium equations.The
structural problems with number of unknowns greater than the
number independent equilibrium equations are called statically
indeterminate.
2
2) Equations based on geometric relations regarding elastic
deformations, produced by the loads.
1) Equilibrium equations based on free body diagram of the structure
or part of the structure.
The following equations are required to solve the problems on
statically indeterminate structure.

COMPOUND BAR
(i) Change in length in all the materials are same.
(ii) Applied load is equal to sum of the loads carried by each bar.
Material(1)
Material(2)
W
L1 L2
3
A compound bar is one which is made of two or more than two
materials rigidly fixed, so that they sustain together an externally
applied load. In such cases

Total load = load carried by material 1
+ load carried by material 2
W = σ1 A1 + σ2 A2 (2)
From Equation (1) & (2) σ1 and σ2 can be calculated
4
E1 L1
E2 L2
(1)σ1 = σ2 × ×
(dL)1 = (dL)2
σ1 .L1 σ2 .L2
E1 E2
=
E1/E2 is called modular ratio

Problems
(1) A load of 300KN is supported by a short concrete
column 250mm square. The column is strengthened by
4 steel bars in corners with total c/s area of 4800mm2
.
If Es=15Ec, find the stress in steel and concrete.
If the stress in concrete not to exceed 4MPa, find the
area of steel required so that the column can support a
load of 600KN.
4
250mm
250mm
Steel

Solution:
Deformation is same
(dL)s = (dL)c
(σs/ Es )× Ls= (σc / Ec)× Lc
σs/ 15Ec= σc/Ec250mm
250mm
Steel
Case(i) : As = 4800mm2
Ac = (250×250) – 4800 = 57,700mm2
σs = 15σc (1)
W = σs As + σc Ac
300 × 103
= 15 σc × 4800 + σc× 57,700
σc = 2.31 N/mm2
2
6

Case (ii)
W= σs As + σc Ac
600× 103
= 15 σc × As + σc Ac
600 × 103
= (15 × 4) As + 4 (250 × 250 – As)
As = 6250 mm2
7

(2) A mild steel rod 5 mm diameter passes centrally through a copper
tube of internal diameter 25mm and thickness 4mm. The
composite section is 600mm long and their ends are rigidly
connected. It is then acted upon by an axial tensile load of 50kN.
Find the stresses & deformation in steel and copper. Take Ecu =
100GPa, Es = 200GPa
8
600mm
Steel
Copper
5mm
25mm
33mm
50KN

Solution:
600mm
Steel
Copper
5mm
25mm
33mm
50KN
Since deformation are same
(dL)s = (dL)cu
(σs / Es)×Ls =( σcu / Ecu )× Lcu
σs / (200 × 103
)= σcu / (100 × 103
)
σs = 2 σcu
50 × 103
= 2σcu ×( π/4) (5)2
+ σcu × π/4 [(33)2
– (25)2
]
(dL)s = [247.72/(200 ×103
)] ×600 0.74mm = (dL)cu
9
W = σs As+ σcu Acu
(dL)s = (σs / Es )× Ls
σcu = 123.86N/mm2
σs = 247.72 N/mm2

(3) Three vertical rods AB, CD, EF are hung from rigid supports and
connected at their ends by a rigid horizontal bar. Rigid bar carries a
vertical load of 20kN. Details of the bar are as follows:
(i) Bar AB :- L=500mm, A=100mm2
, E=200GPa
(ii) Bar CD:- L=900mm, A=300mm2
, E=100GPa
(iii) Bar EF:- L=600mm, A=200mm2
, E=200kN/mm2
If the rigid bar remains horizontal even after loading, determine the
stress and elongation in each bar.
Solution:
600mm
900mm500mm
A
B D E
C
F
20kN
10

W = (σAB×AAB)+ (σCD×ACD)+ (σEF×AEF)
20 × 103
= (3.6 × σCD× 100) + (σCD× 300) + (3σCD × 200)
σCD = 15.87N/mm2
σAB = 3.6 × 15.87 = 57.14N/mm2
σEF = 3 × 15.87 = 47.61N/mm2
dLAB = (σAB/ EAB)× LAB= [57.14/(200 × 103
)] ×500
Deformations are same
(dL)AB = (dL)CD = (dL)EF
(σAB / EAB) × LAB = (σCD / ECD)× LCD = (σEF / EEF) × LEF
[σAB/(200 ×103
)]× 500 =[σCD /(100 ×103
)] ×900= [σEF /(200× 103
)]× 600
σAB = 3.6× σCD ; σEF = 3× σCD
11

(4) Two copper rods and one steel rod together supports as shown
in figure. The stress in copper and steel not to exceed 60MPa
and 120MPa respectively. Find the safe load that can be
supported. Take Es = 2Ecu
W
Copper
(30mm×30mm)
Copper
(30mm×30mm)
Steel
(40mm×40mm)
120mm
80mm
12

Solution:
Deformations are same i.e., (dL)s = (dL)cu
(σs / Es) × Ls = (σcu / Ecu )× Lcu
(σs / 2Ecu )× 200 =( σcu / Ecu )× 120 => σ s = 1.2 σcu
Let σcu=60MPa=60N/mm2
,
σs=1.2x60 = 72N/mm2
< 120N/mm2
(safe)
Safe load = W = σs× As + 2( σcu ×Acu )
= 72(40× 40) + 2 ×[60×(30 × 30)]
Safe load = W = 223.2 × 103
N = 223.2 kN
13

(5)A rigid bar AB 9m long is suspended by two vertical rods at its
end A and B and hangs in horizontal position by its own weight.
The rod at A is brass, 3m long, 1000mm2
c/s and Eb = 105
N/mm2
.
The rod at B is steel, length 5m, 445mm2
c/s and Es = 200GPa. At
what distance x from A, if a vertical load P = 3000N be applied if
the bar remains horizontal after the load is applied.
9m
5m
3m
Steel
A B
3000N
Brass
x
14

W= σsAs + σbAb
3000= (1.2 σb × 445) + (σb × 1000)
σb = 1.95N/mm2
σs = 2.34N/mm2
Deformations are same i.e., (dL)b = (dL)s
(σb / Eb )× Lb = (σs / Es )× Ls
(σb / 105
) ×(3× 103
) = [σs / (200 × 103
)] × [5 × 103
]
=> σs = 1.2 σb
+ve ΣMA= 0 => -(3000) (x) + (2.34 × 445) × 9000 = 0
15
x = 3123.9 mm = 3.12m from A
If the load of 3000N is kept at a distance of 3.12m from A, bar AB will
remain horizontal.

(6) A mild steel bar of c/s 490mm2
is surrounded by a copper tube of
210mm2
as shown. When they are placed centrally over a rigid bar,
it is found that steel bar is 0.15 mm longer. Over this unit a rigid
plate carrying a load of 80 kN is placed. Find the stress in each bar,
if the length of the compound bar is 1m. Take Es = 200 GPa, Ecu =
100 GPa.
16
Steel bar
80kN
Copper tube
0.15mm
1000mm

(dL)s = (dL)cu + 0.15
(σs / Es ) × Ls =( σcu / Ecu )× Lcu + 0.15
[σs / (200 × 103
)] × 1000.15 = {[σcu / (100 × 103
)] ×1000} + 0.15
σ s = 2 σcu + 30
W = σs ×As + σcu ×Acu
σcu = 54.87N/mm2 σs = (2×54.87) + 30 = 139.84N/mm2
Solution: 17
80× 103
= [( 2σcu + 30)× 490] + (σcu × 210)

Temperature Stress
Any material is capable of expanding or contracting freely due to rise
or fall in temperature. If it is subjected to rise in temperature of T˚C,
it expands freely by an amount ‘αTL’ as shown in figure. Where α is
the coefficient of linear expansion, T˚C = rise in temperature and L =
original length.
L
A B
L
A
B
L
A B
B´
P
αTL
18

From the above figure it is seen that ‘B’ shifts to B' by an amount
‘αTL’. If this expansion is to be prevented a compressive force is
required at B'.
Temperature strain = αTL/(L + αTL) ≈ αTL/L= αT
Hence the temperature strain is the ratio of expansion or contraction
prevented to its original length.
If a gap δ is provided for expansion then
19
Temperature stress = αTE
Temperature stress = [(αTL – δ)/L] E
Temperature strain = (αTL – δ) / L

Temperature stress in compound bars:-
Material(2)
Material(1)
α2TL
∆
α1TL
(dL)1
P1
(dL)2
P2
x
x
When a compound bar is subjected to change in temperature, both the
materials will experience stresses of opposite nature.
20
σ1A1 = σ2A2 (there is no external load)
σ1 = ( σ2A2)/A1 (1)
Compressive force on material (1) = tensile force on material (2)

As the two bars are connected together, the actual position of the bars
will be at XX.
Actual expansion in material (1) = actual expansion in material (2)
α1TL – (dL)1 = α2TL + (dL)2
α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L
αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2)
From (1) and (2) magnitude of σ1 and σ2 can be found out.
21

(1) A steel rail 30m long is at a temperature of 24˚C. Estimate the
elongation when temperature increases to 44˚C. (1) Calculate
the thermal stress in the rail under the following two conditions :
(i) No expansion gap provided
(ii) If a 6mm gap is provided for expansion
(2) If the stress developed is 60MPa , what is the gap left
between the rails?
Take E = 200GPa, α = 18 x 10-6
/˚C
PROBLEMS 22

Free expansion αTL = 18 × 10-6
×(44-24) × 30 × 103
= 10.8mm
i) No expansion joints provided:-
Temperature stress = αTE
= 18 × 10-6
× 20× 200 × 103
= 72N/mm2
ii) 6mm gap is provided for expansion
Temperature stress = [( αTL – δ) / L] E
= [(10.8 – 6)/(30 × 103
)] ×200 × 103
= 32N/mm2
When stress = 60MPa
Temperature stress = [( αTL – δ) / L] E
= [(10.8 – δ ) / (30 × 103
)] × 200 × 103
Solution:
δ = 1.8mm
23

(2) A steel bar is placed between two copper bars. Steel bar and
copper bar has c/s 60mm × 10mm and 40mm × 5mm
respectively connected rigidly on each side. If the temperature
is raised by 80°C, find stress in each metal and change in length.
The length of bar at normal temperature is 1m. Es = 200GPa,
Ecu= 100GPa, αs = 12 x 10-6
/° C, αcu = 17x10-6
/ ° C
Steel
Copper
αcuTLcu
(dL)cu
Pcu
(dL)s
Ps
Copper
40mm
60mm
40mm
1000mm
Pcu
x
αsTLs
x∆
Solution
:
24

Compressive force on copper bar = tensile force on steel bar
2σcu ×Acu = σs ×As
2σcu ( 40 × 5) = σs ( 60 × 10)
αcuTLcu - (dL)cu = αsTLs + (dL)s
αcuTLcu - (σcu / Ecu) Lcu = αsTLs + (σs / Es) Ls
=> σcu = 1.5σs
Actual expansion in copper = Actual expansion in steel
Since Lcu = Ls
(17 × 10-6
× 80 )– 1.5σs /(100 × 103
) = (12× 10-6
× 80) + σs /(200 × 103
)
σs = 20N/mm2
(T) σcu=1.5 × 20 = 30N/mm2
(C)
Δ = Change in length = αcu ×T×Lcu – (σcu / Ecu) Lcu
= 17×10-6
× 80 × 1000 – [30/(100 × 103
)]× 1000
Δ = 1.06mm
25

(3) A horizontal rigid bar weighing 200 kN is hung by three vertical
rods each of 1m length and 500mm2
c/s symmetrically as shown.
Central rod is steel and the outer rods are copper. Temperature rise is
40ºC. (1) Determine the load carried by each rod and by how much the
horizontal bar descend? Given Es = 200GPa. Ecu=100GPa. αs =1.2 x 10-
5
/ºC. αcu=1.8x 10-5
/ºC. (2) What should be the temperature rise if the
entire load of 200kN is to be carried by steel alone.
(dL)T
(dL)L
(dL)T
(dL)L
(dL)T
(dL)L
PcuPsPcu
Copper CopperSteel
200kN
26

[ (dL) T +(dL) L]cu = [ (dL)T +(dL)L]st
[αcuTLcu +( PcuLcu / Acu Ecu ) ] = [αsTLs +( PsLs / AsEs )]-----(1)
For equilibrium condition ∑Fv = 0 => Ps + 2Pcu = 200 × 103
Ps = 200 × 103
– 2Pcu
Substituting in (1) [1.8 × 10-5
× 40 + Pcu/ (500 × 100 × 103
)]
={ 1.2 × 10-5
× 40 +[ (200 × 103
– 2Pcu ) / (500 × 200 × 103
)]}
Pcu = 44,000N Ps = 112 × 103
NPs = 200 × 103
– 2 × 44,000
Elongation = αcuTLcu + (Pcu ×Lcu )/(Acu ×Ecu)
= 1.8 ×10-5
× 40 ×1000 +[ 44000 × 1000/(500 × 100 × 103
)]
dL=1.6mm
27

Case (ii)
[ (dL)T ]cu = [ (dL)L +(dL)T]s
=> αcuTLcu = PsLs / AsEs +αsTLs
=> [1.8 × 10-5
×T] = [(200 × 103
) / (500 × 200 × 103
) +1.2 × 10-5
× T]
=> T = 333.33ºC
PcuPsPcu
(dL)T
(dL)L
(dL)T
28
CopperSteel
(dL)T 200kN
Copper

(4) A rigid bar AB is hinged at A and is supported by copper and steel
bars as shown each having c/s area 500mm2
. If temperature is raised
by 50ºC, find stresses in each bar. Assume Ecu = 100 Gpa. Es=
200GPa, αs = 1.2 x 10-5
/ºC αcu = 18 x 10-6
/ºC
C B
B'
αsTLs
Ps
(dL)s
(dL)cu
C'
αcuTLcu
Pcu
C"
B"
RA
A
29
Copper 200mm
D
Steel
150mm
E
A C
B
200mm 300mm

Copper:- Temp.Stress = [(αcuTLcu)– (dL)cu / 200] Ecu
Pcu= {[(αcuTLcu)– (dL)cu ]/ 200}× (Ecu × Acu)
= [(18 × 10-6
× 50 × 200) – (dL)cu)/200] 100 ×103
× 500
Pcu= [45 × 103
- 250 ×103
(dL)cu]-----------(1)
Steel:- Ps = {[(dL)s – αsTLs] / 150} EsAs
Ps = {[(dL)s – 12 x 10-6
× 50 × 150]/150 ]} × (200× 103
× 500)
Ps = 6.66 ×105
(dL)s – 6 × 104
----------------(2)
From similar Δle (dL)cu/200 = (dL)s/500 => (dL)s = 2.5(dL)cu --- (3)
Pcu = 2.5Ps (4)
30
+ ve Σ MA = 0 => -Ps × 500 +Pcu × 200 = 0

Substituting (1) & (2) in (4)
(45 × 103
)– [250 ×103
(dL)cu]= 2.5 {[ 6.66× 105
(dL)s] – (6×104
)}
45 ×103
– 250 ×103
(dL)cu= 2.5 {[ 6.66 × 105
× 2.5(dL)cu]– (6×104
)}
(dL)cu= 0.044mm and (dL)s=0.11mm
Substituting in (1) & (2)
Pcu = (45 × 103
) – (250 × 103
× 0.044) = 34,000N
Ps = 34,000/2.5 = 13,600N
σs = Ps / As =13,600/500 = 27.2N/mm2
(T)
σcu = Pcu / Acu = 34,000/500 = 68N/mm2
(C)
31

(5) A composite bar is rigidly fixed at A and B.Determine the
reaction at the support when the temperature is raised by 20ºC. Take
EAl = 70GPa, Es = 200GPa, αAl = 11 x 10-6
/ºC, αs = 12 x 10-6
/ºC.
A = 600mm2
A = 300mm2
40kN
Aluminium
1m
Steel
3m
B
A
32
A = 600mm2
A = 300mm2
40kN
AL
1m
Steel
3m
RB
RA
Solution:

RA + RB = 40 ×103
=> RA = 40× 103
– RB
[(dL)T + (dL)L ]al + [ (dL)T + (dL)L ]s = 0
{[(α T L)– [(RA x L) /(A × E )]}al+ {[(αTL)+[( RB×L)/(A×E)]s= 0
Aluminium:-
RA RA
Steel:-
RB RB
33
40kN

(11 × 10-6
× 20 × 1000) –[{(40 × 103
– RB)×1000}/(600 × 70 × 103
)] + (12
×10-6
× 20 ×3000 )+ (RB × 3000)/(300 × 200 × 103
)= 0
0.22 – 0.95 + (2.38 × 10-5
RB) + 0.072 +( 5 ×10-5
RB )= 0
7.38 × 10-5
RB = 0.658
34
RA = 8915.99 ≈ 8916N ( )
RB = 31,084N ( )

(6) A bar is composed of 3 segments as shown in figure. Find the
stress developed in each material when the temperature is raised by
50˚C under two conditions
i)Supports are perfectly rigid
ii) Right hand support yields by 0.2mm
Take Es = 200GPa, Ecu =100GPa, Eal = 70GPa, αs = 12 x 10-6
/ºC,
αcu = 18 x 10-6
/ºC, αal = 24 x 10-6
/ºC.
A=200mm2
A=400mm2
A=600mm2
150mm
200mm
150mm
Steel Copper Aluminium
35

Case(i): Supports are perfectly rigid
(dL)s + (dL)cu + (dL)al = αsTLs + αcuTLcu + αalTLal
= (12× 10-6
×50 × 150 ) +(18 × 10 -6
× 50 × 200) +(24 × 10-6
× 50 ×150)
= 0.45mm
(σs/Es) Ls + (σcu/Ecu) Lcu + (σal/Eal) Lal = 0.45mm
[σs/(200× 103
)]×150+[σcu/(100×103
)]×200+[σal/(70× 103
)]×150=0.45 -(1)
From principle of compound bars
σsAs = σcuAcu = σalAal => σs × 200 = σcu × 400 = σal × 600
σs=2σcu σal = 0.67σcu
36

Substituting in (1)
[2σcu/(200 × 103
)]×150+[σcu/(100 × 103
)]×200+[0.67σcu/(70×103
)]
×150=0.45
σcu = 91.27N/mm2
σs = 2σcu = 182.54N/mm2
, σal = 0.67 σcu = 61.15N/mm2
Case (ii) Right hand support yield by 0.2mm
(σs/Es) Ls + (σcu/Ecu) Lcu + (σal/Eal) Lal = 0.45 – 0.2=0.25
[2σcu/(200 × 103
)]×150 + [σcu/(100 × 103
)]×200
+ [0.67σcu/(70×103
)] ×150 = 0.25
σcu = 50.61N/mm2
σs = 2σcu = 101.22N/mm2
; σal = 0.67 σcu = 33.91N/mm2
37

Exercise problems 38
1) A circular concrete pillar consists of six steel rods of 22mm diameter
each reinforced into it. Determine the diameter of pillar required when it
has to carry a load of 1000kN. Take allowable stresses for steel &
concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15
ANS: D=344.3mm

39
2) Determine the stresses & deformation induced in Bronze & steel as
shown in figure. Given As=1000mm2
, Ab=600mm2
, Es= 200Gpa, Eb=
83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)
160kN
Bronze Bronze
Steel

40
3) A cart wheel of 1.2m diameter is to be provided with steel tyre.
Assume the wheel to be rigid. If the stress in steel does not exceed
140MPa, calculate minimum diameter of steel tyre & minimum
temperature to which it should be heated before on to the wheel.
ANS: d=1199.16mm T=58.330
C
4) A brass rod 20mm diameter enclosed in a steel tube of 25mm internal
diameter & 10mm thick. The bar & the tube are initially 2m long &
rigidly fastened at both the ends. The temperature is raised from 200
C to
800C. Find the stresses in both the materials.
If the composite bar is then subjected to an axial pull of 50kN, find the
total stress. Es=200GPa, Eb=80GPa, αs=12×10-6
/0
C, αb=19×10-6
/0
C.
ANS: σb=8.81N/mm2
( C ) , σs=47.99N/mm2
( T )

temperature stresses in Strength of materials

temperature stresses in Strength of materials

More Related Content

What's hot

Similar to temperature stresses in Strength of materials

More from Shivendra Nandan

Recently uploaded

In this document

temperature stresses in Strength of materials