statically indeterminate problems problems involving...

Statically Indeterminate Problemsand

Problems Involving Two Materials(Strength of Materials)

Dave Morgan<[email protected]>

Statically Indeterminate Problems and Problems Involving Two Materials – p. 1/30

Statically Indeterminate Problems

A

B

C

30 kN

10 kN

We have used the laws of statics to analyse

problems such as the one illustrated:



A

B

C

30 kN

10 kN

20 kN



ΣFy = 0, so there is a reaction force of

20 kN at A



A

B

C

30 kN

10 kN

20 kN




20 kN at A

To find the internal forces in the segment

BC:



A

B

C

30 kN

10 kN

20 kN

M M




20 kN at A


BC:Insert a section M-M through

segment BC



A

B

C

30 kN

10 kN

20 kN

M M




20 kN at A



segment BCConsider only the segment from C to

M-M



A

B

C

30 kN

10 kN

20 kN

M M

30 kN




20 kN at A




M-MΣFy = 0, so there is an internal force

of 30 kN at M



A

B

C

30 kN

10 kN

20 kN

M M

30 kN




20 kN at A




M-MΣFy = 0, so there is an internal force

of 30 kN at MTBC = 30 kN (tension)



A

B

C

30 kN

10 kN

20 kNSimilarly, we can find the internal force

within segment AB:



A

B

C

30 kN

10 kN

20 kN

N N

Similarly, we can find the internal force

within segment AB:

Insert a section N-N through segment

AB



A

B

C

30 kN

10 kN

20 kN

N N


within segment AB:


AB

Consider only the segment from A to

N-N



A

B

C

30 kN

10 kN

20 kN

N N

20 kN


within segment AB:


AB


N-N

ΣFy = 0, so there is an internal force

of 20 kN at N-N



A

B

C

30 kN

10 kN

20 kN

N N

20 kN


within segment AB:


AB


N-N

ΣFy = 0, so there is an internal force

of 20 kN at N-N

TAB = 20 kN (tension)



Structures where forces can be determined using thestatic equilibrium equations alone (ΣFx = 0,ΣFy = 0 andΣMA = 0) are calledstaticallydeterminate structures. The previous example is astatically determinate structure.




Structures where the forces cannot be determined inthis way are calledstatically indeterminatestructures.




Structures where the forces cannot be determined inthis way are calledstatically indeterminatestructures.

Statically indeterminate structures are often analysedusing the conditions of axial deformation given by

δ =PL

AEStatically Indeterminate Problems and Problems Involving Two Materials – p. 4/30


Example: Consider a bar AB supported at bothends by fixed supports, with an axial force of 12 kNapplied at C as illustrated. Find the reactions at thewalls

A C B

500 mm 400 mm

12 kN




A C B

500 mm 400 mm

12 kN

Solution: First, draw a free body diagram:




A C B

500 mm 400 mm

12 kN

Solution: First, draw a free body diagram:

A C B

500 mm 400 mm

12 kNRA RB



A C B

500 mm 400 mm

12 kNRA RB



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12

Two unknowns and a single equation; the problem isstatically indeterminate.



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12

Two unknowns and a single equation; the problem isstatically indeterminate.The supports at A and B are fixed soδAB = 0.



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12


⇒ δAC + δCB = 0



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12


⇒ δAC + δCB = 0

⇒−RA×500

AE+ RB×400

AE= 0



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12


⇒ δAC + δCB = 0

⇒−RA×500

AE+ RB×400

AE= 0

⇒ 400RB = 500RA



A C B

500 mm 400 mm

12 kNRA RB

ΣFx = RA + RB − 12 = 0

⇒ RA + RB = 12


⇒ δAC + δCB = 0

⇒−RA×500

AE+ RB×400

AE= 0

⇒ 400RB = 500RA

Now we have two equations and two unknowns; we cansolve forRA andRB



RA + RB = 12

400RB = 500RA



RA + RB = 12

400RB = 500RA

⇒ RA = 12 − RB



RA + RB = 12

400RB = 500RA

⇒ RA = 12 − RB

⇒ 400RB = 500 (12 − RB)



RA + RB = 12

400RB = 500RA

⇒ RA = 12 − RB

⇒ 400RB = 500 (12 − RB)

⇒ 900RB = 6000



RA + RB = 12

400RB = 500RA

⇒ RA = 12 − RB

⇒ 400RB = 500 (12 − RB)

⇒ 900RB = 6000

⇒ RB = 6.667 kN



RA + RB = 12

400RB = 500RA

⇒ RA = 12 − RB

⇒ 400RB = 500 (12 − RB)

⇒ 900RB = 6000

⇒ RB = 6.667 kN

⇒ RA = 5.333 kN



A C B

500 mm 400 mm

12 kN5.333 kN 6.667 kN



Exercise: FindRA andRB for the problem illustrated:

A C B

a b

P




A C B

a b

P

Solution: Draw free body diagram




A C B

a b

P

Solution: Draw free body diagram

A C B

a b

RA RBP



A C B

a b

RA RBP

Solution:



A C B

a b

RA RBP

Solution: RA + RB − P = 0



A C B

a b

RA RBP


⇒ RA = P − RB



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0

⇒ aRA = bRB



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0

⇒ aRA = bRB

⇒ a (P − RB) = bRB



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0

⇒ aRA = bRB

⇒ a (P − RB) = bRB

⇒ aP = (a + b)RB



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0

⇒ aRA = bRB

⇒ a (P − RB) = bRB

⇒ aP = (a + b)RB

⇒ RB =(

aa+b

)

P



A C B

a b

RA RBP


⇒ RA = P − RB

δAC + δCB = 0

⇒RA×aAE

+ −RB×bAE

= 0

⇒ aRA = bRB

⇒ a (P − RB) = bRB

⇒ aP = (a + b)RB

⇒ RB =(

aa+b

)

P

⇒ RA =(

ba+b

)

PStatically Indeterminate Problems and Problems Involving Two Materials – p. 10/30

Problems Involving Two Materials

Steel-reinforced concrete is used in the construction ofmany structures:

Bridges

Basements

High-Rise Buildings

Stadia, such as the SaddleDome or theSpeed-Skating Oval



Concrete has a high load-bearing capacity incompression but is not very strong under a tensileload.




Steel rod has high load-bearing capacity in tensionbut buckles easily under compression.




Steel rod has high load-bearing capacity in tensionbut buckles easily under compression.

Combining steel rod and concrete gives a buildingmaterial with both good tensile and compressiveload-bearing qualities.



Steel in a concrete column also helps the concrete’scompressive strength:




When a column is loaded, it deforms(

δ = PLAE

)





δ = PLAE

)

Under compression,δ is negative and there isnegative axial strain





δ = PLAE

)

Under compression,δ is negative and there isnegative axial strainConsequently, there is a positive transverse strain(ǫt = −µǫa)





δ = PLAE

)


The concrete is under tension laterally





δ = PLAE

)


The concrete is under tension laterallyHorizontal steel-reinforcing increases the lateraltensile strength of the column



A concrete footing ispoured:

It contains steel rebarthroughout

Steel extrudes fromthe top of the footing

This will be attachedto the steel for thecolumn.



Steel is tied for the column



A frame is built around thesteel and the concretecolumn is poured



We can useδC = PC ·LC

AC ·ECto calculate the deformation

of concrete under a load






We can useδS = PS ·LS

AS ·ESto calculate the deformation

of steel under a load









How can we calculate the deformation of asteel-reinforced concrete column?










EC is not the same asES so we cannot simplyapplyδ = PL

AEfor the whole column










EC is not the same asES so we cannot simplyapplyδ = PL

AEfor the whole column

We cannot solve this problem directly using theequations of statics, so this is astatically-indeterminate problem



Example: A concrete column has a

diameter of300 mm. The column has6 steel

reinforcing rods.

Each rod has a cross-sectional area of

200 mm2. (See plan view)

ES = 210 GPa andEC = 25 GPa

The column is1.15 m long and has a load of

1.37 MN is applied to a rigid steel plate at

the top of the column (the plate distributes

the load evenly over the top of the column).

1.37 MN

1.15 m

A = 200 mm2

D = 300 mm



Example: A concrete column has a

diameter of300 mm. The column has6 steel

reinforcing rods.

Each rod has a cross-sectional area of

200 mm2. (See plan view)

ES = 210 GPa andEC = 25 GPa

The column is1.15 m long and has a load of

1.37 MN is applied to a rigid steel plate at

the top of the column (the plate distributes

the load evenly over the top of the column).

Find the stress in the steel and in the

concrete, and the deformation under the load

1.37 MN

1.15 m

A = 200 mm2

D = 300 mm



Solution: Let PS be the total reaction force of the

six steel rods andPC the reaction force of the

concrete.

PC

PS

6

PS

6

1370 kN





concrete.

ΣFy = PS + PC − 1370 = 0

PC

PS

6

PS

6

1370 kN





concrete.

ΣFy = PS + PC − 1370 = 0

PS + PC = 1370 kN

PC

PS

6

PS

6

1370 kN





concrete.

ΣFy = PS + PC − 1370 = 0

PS + PC = 1370 kN

We have a single equation with two unknowns, so

the problem is statically indeterminate.

PC

PS

6

PS

6

1370 kN





concrete.

ΣFy = PS + PC − 1370 = 0

PS + PC = 1370 kN



The concrete and the steel rods deform (contract)

by the same amount,δ, so...

PC

PS

6

PS

6

1370 kN





concrete.

ΣFy = PS + PC − 1370 = 0

PS + PC = 1370 kN



The concrete and the steel rods deform (contract)

by the same amount,δ, so...

PS · LS

AS · ES

= δ =PC · LC

AC · EC

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

⇒PS×1150

1200×(200×103)= PC×1150

“

π×3002

4−1200

”

×(25×103)

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

⇒PS×1150

1200×(200×103)= PC×1150

“

π×3002

4−1200

”

×(25×103)

⇒PS

1200×200 = PC

69486×25PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

⇒PS×1150

1200×(200×103)= PC×1150

“

π×3002

4−1200

”

×(25×103)

⇒PS

1200×200 = PC

69486×25

⇒ PS = 1200×20069486×25 · PC

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

⇒PS×1150

1200×(200×103)= PC×1150

“

π×3002

4−1200

”

×(25×103)

⇒PS

1200×200 = PC

69486×25

⇒ PS = 1200×20069486×25 · PC

⇒ PS = 0.13816PC

PC

PS

6

PS

6

1370 kN



Solution:PS ·LS

AS ·ES= PC ·LC

AC ·EC

⇒PS×1150

(6×200)×ES= PC×1150

“

π×3002

4−(6×200)

”

×EC

⇒PS×1150

1200×(200×103)= PC×1150

“

π×3002

4−1200

”

×(25×103)

⇒PS

1200×200 = PC

69486×25

⇒ PS = 1200×20069486×25 · PC

⇒ PS = 0.13816PC

We now have two equations for the two unknowns,

PS andPC .

PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

⇒ 0.13816PC + PC = 1370

PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

⇒ 0.13816PC + PC = 1370

⇒ PC = 13701+0.13816

PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

⇒ 0.13816PC + PC = 1370

⇒ PC = 13701+0.13816

⇒ PC = 1203.7 kN

PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

⇒ 0.13816PC + PC = 1370

⇒ PC = 13701+0.13816

⇒ PC = 1203.7 kN

⇒ PS = 166.3 kN PC

PS

6

PS

6

1370 kN



Solution:

PS + PC = 1370

PS = 0.13816PC

⇒ 0.13816PC + PC = 1370

⇒ PC = 13701+0.13816

⇒ PC = 1203.7 kN

⇒ PS = 166.3 kN

We can now calculate the stress in the steel and in

the concrete

PC

PS

6

PS

6

1370 kN



Solution: Find the stress in the concrete:

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa

Find the stress in the steel:

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa


PS = 152 kN PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa


PS = 152 kN

⇒ σS = 152(6×200)

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa


PS = 152 kN

⇒ σS = 152(6×200)

⇒ σS = 0.1267 kN

mm2

PC

PS

6

PS

6

1370 kN




PC = 1098 kN

⇒ σC = PC

A

⇒ σC = 1098π×3002

4−(6×200)

⇒ σC = 0.0580 kN

mm2

⇒ σC = 58.0 MPa


PS = 152 kN

⇒ σS = 152(6×200)

⇒ σS = 0.1267 kN

mm2

⇒ σS = 126.7 MPa

PC

PS

6

PS

6

1370 kN



Solution: Find the deformation in the concrete:




δC = PC ·LC

AC ·EC




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)

⇒ δC = 0.00221 mm




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)

⇒ δC = 0.00221 mm

Find the deformation in the steel (if we’ve done our calculations correctly,

thenδS = δC):

δS = PS ·LS

AS ·ES




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)

⇒ δC = 0.00221 mm


thenδS = δC):

δS = PS ·LS

AS ·ES

⇒ δS =152×(3.5×103)1200×(200×103)




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)

⇒ δC = 0.00221 mm


thenδS = δC):

δS = PS ·LS

AS ·ES

⇒ δS =152×(3.5×103)1200×(200×103)

⇒ δS = 0.00222 mm




δC = PC ·LC

AC ·EC

⇒ δC =1098×(3.5×103)

(π×3002

4−1200)×(25×103)

⇒ δC = 0.00221 mm


thenδS = δC):

δS = PS ·LS

AS ·ES

⇒ δS =152×(3.5×103)1200×(200×103)

⇒ δS = 0.00222 mm

The small difference in deformation is due to rounding errors



Exercise: A hollow square steel

structural section has outside dimensions of

115 mm× 115 mm and inside dimensions of

105 mm× 105 mm. It is filled with concrete,

as shown in plan view (upper right). The

section is3.5 m and supports a compressive

load of250 kN.

ES = 200 GPa andEC = 20 GPa.

FindσS , σC andδ

105 mm

115 mm

250 kN

175 mm



Solution: Find the areas of the steel and

of the concrete:

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

AC = 105 × 105

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

AC = 105 × 105

= 11025 mm2

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

AC = 105 × 105

= 11025 mm2

Let PS be the reaction force of the steel and

PC the reaction force of the concrete.

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

AC = 105 × 105

= 11025 mm2


PC the reaction force of the concrete. Then,

ΣFy = PS + PC − 250 = 0

105 mm

115 mm

250 kN

175 mm




of the concrete:

AS = (2 × 115 × 5) + (2 × 105 × 5)

= 2200 mm2

AC = 105 × 105

= 11025 mm2


PC the reaction force of the concrete. Then,

ΣFy = PS + PC − 250 = 0

PS + PC = 250 kN

105 mm

115 mm

250 kN

175 mm



Solution: The steel casing and the

concrete both deform by the same amount

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

⇒ PC = 11025×202200×200 · PS

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

⇒ PC = 11025×202200×200 · PS

⇒ PC = 0.5011PS

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

⇒ PC = 11025×202200×200 · PS

⇒ PC = 0.5011PS

ΣFy = 0 so

PC + PS − 250 = 0

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

⇒ PC = 11025×202200×200 · PS

⇒ PC = 0.5011PS

ΣFy = 0 so

PC + PS − 250 = 0

⇒ PC = 250 − PS

105 mm

115 mm

250 kN

175 mm





PC ·LC

AC ·EC= PS ·LS

AS ·ES

⇒(PC×103)×175

11025×(200×103)=

(PS×103)×175

2200×(200×103)

⇒PC

11025×20 = PS

2200×200

⇒ PC = 11025×202200×200 · PS

⇒ PC = 0.5011PS

ΣFy = 0 so

PC + PS − 250 = 0

⇒ PC = 250 − PS

Now we have two equations for the two

unknowns,PC andPS

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

⇒ PC = 83.5 kN

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

⇒ PC = 83.5 kN

Now, find the stress in the steel:

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

⇒ PC = 83.5 kN


σS = PS

AS

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

⇒ PC = 83.5 kN


σS = PS

AS

⇒ σS = 166.5×103

2200

105 mm

115 mm

250 kN

175 mm



Solution:

PC = 0.5011PS

PC = 250 − PS

⇒ 250 − PS = 0.5011PS

⇒ PS = 2501+0.5011

⇒ PS = 166.5 kN

⇒ PC = 83.5 kN


σS = PS

AS

⇒ σS = 166.5×103

2200

⇒ σS = 75.7 MPa

105 mm

115 mm

250 kN

175 mm



Solution: Now, find the stress in the

concrete:

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

⇒ σC = 7.57 MPa

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

⇒ σC = 7.57 MPa

Now, find the deformation:

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

⇒ σC = 7.57 MPa


δ = PC ·LC

AC ·EC

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

⇒ σC = 7.57 MPa


δ = PC ·LC

AC ·EC

⇒ δ =(83.5×103)×175

11025×(20×103)

105 mm

115 mm

250 kN

175 mm




concrete:

σC = PC

AC

⇒ σC = 83.5×103

11025

⇒ σC = 7.57 MPa


δ = PC ·LC

AC ·EC

⇒ δ =(83.5×103)×175

11025×(20×103)

⇒ δ = 0.0663 mm

105 mm

115 mm

250 kN

175 mm


Statically Indeterminate Problems and Problems Involving Two Materials

Created by Dave Morgan using LATEX 2εandProsper on January 26, 2006


statically indeterminate problems problems involving...

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