1/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

STATICALLY DETERMINED PLANE BAR STURCTURES (FRAMES, ARCHES)

2/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Formal definition: A frame is a plane (2D) set of beams connected at stiff and/or hinged joints (corners)

Joints have to be in the equilibrium!

X = 0Y = 0MK = 0

?

? !?!

!?!

Frames

STIFF JOINT HINGED JOINT

3/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

What is the difference between beams and frames?

Why do we need to make frames?

Beam FrameBeam or frame?

Hey, you!

Frames

4/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Equilibrium equationsX = 0Y = 0MK = 0

MA2 = 0MA1 = 0

MAn = 0…………

+ For n hinged joints (if any!) at A1, A2 …An points

+ kinematic stability of a structure (c.f. Theoretical Mechanics)

Centre of instability

Examples of unstable structures

HYPER-STIFF

UNSTABLE

Frames

5/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

After we determine reactions and check stability we can deal with a frame as a set of individual beams, applying all techniques which have been demonstrated for beams. But, besides of diagrams of M and Q we have to make diagrams of N, too.

Some problems can be encountered with sloping members

x

y

Wq’=q·Δx/Δs=q ·cosW = q·Δx

W

Δs

Δx

x

s y

q’

qs =q’·cos= q·cos2Δx/Δs = cosq

q s

s ns =q’·sin= q·sin ·cos

= q’·Δs

qsq’

ns

Frames

n s

6/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Example: Diagrams of M, Q, N for a simple frame1,5 kN/m

2 kN

2 m

1,5 m

1 m 2 kN1 kN

2 kN

sin=0,6 cos =0,8

+-

-

+

Q [kN]

M [kNm]

0,8

31

3

0,33

1

2

1,62

1,2

0,6

2 2

0,5 m

+-

-

2N [kN]

2

Frames

Q

No

n

3kN

7/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

2

Checking the equilibrium at a joint

M [kNm]

31

3

0,25

1

+-

2

1,2

0,6

-

2N [kN]

-

+

Q [kN]

0,8

1,62

2 2

0,5 m

+-

2

3

3

3

3 2

3

3 2

2

2

Q

No

nFrames

8/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

FRAMEARCH

M

N

+ --

Stones and other brittle materials do not sustain an extension

Arches

Formal definition: An arch is a plane (2D) set of curved beams connected at stiff and/or hinged joints

9/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

X = 0Y = 0MK = 0

Mc = 0

C

Arches

10/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

x

y

2l

xC

x

yC

CC

y

Parabolic arch

Semi-circular arch

To determine reactions we only need to know position and magnitude of loads and position of the hinge and supports

But to determine the cross-sectional forces we do need the equation describing shape of the arch: including coordinates of any point and its tangent.

y = a + bx + cx2x = r·cos

rC

rC

xC , yC r, C

h

a,b,c from: for x = 0 y = 0

for x = l y = hfor x = 2l y = 0

(Symetric arch)

nn

= arctg dy/dx

Parabolic arch

Semi-circular arch

(in polar coordinates)

Arches

11/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Example: parabolic arch under concentrated force

x

y

l l

h

C

A B

Mc = 0

RA RBVB

HB

X = 0

MA = 0 P·l - VB ·2l =0

VA

HA

VB = P/2

HA/VA = HB/VB = l/h

Higher the ratio l/h (i.e. lower the ratio h / l) – higher the value of horizontal reaction H

HA = HB = VA· l/h

P

Y = 0 = VA

Arches

12/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Arches

x

y

l l

h

C

A B

RA

VBVA = P/2

HA=

(P/2) ·(l/h)

P

Symmetry axis

RA RBM

Symmetry axis

RB

Δ

M = RA·Δ

ΔR

A

N Q

NA

QA

NC=HAQA = VA

13/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

VA=P/2

HA=(P/2)(l/h)

QV

QH NV

NH Bar axis

sincos AAHV HVQQQ

QN

nnN

Q

cossin AAHV HVNNN cossin2 hlPN

A

CAt C: =0 2PQC

hlPNC 2

At A: ǂ0

For (l/h)>>1 – shallow arch

For (l/h)<<1 – steep arch

0sin lhtg 1cos

0cos12sin112sin12 PtgPhlPQA

hlPN A 02

0cos,1sin2

sincos2 hlPQ

000202 PhlPQA

2012 PPN A

hl

AN

14/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

N Q

|NA|>P/2

NC=HA =- (P/2) (l/h)

+

- --

+

QA<P/2

QA = VA=P/2

Anti-symmetric axis

P/2

P/2

Symmetry axis

Q

No

n

15/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

P

r

r r

P/2

P/2

P/2

P/2

M 0,3r

0,2Pr

P/2

P/2

P/2

P/2

P/2

P/2

P/2

Q

P/2

P/2

P/2

P/2

P/2 +

- P/2

P/2+

-P/2

0,7P

N

Q

N

M

o

n

tension

+-+

Example: semi-circular arch under horizontal force

16/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

r

P

r r

r

P

r r

r

P

r

r r

r

P

r

r r

Quantitative comparison of frame, quasi-arch and arch

17/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

r r

P

P

r

P/2P/2

M

Q N

+-

+

Pr

Pr/2

Pr/2

P/2

P

P

Q

N

M

o

n

„Frame”

18/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

P

P/2r r

P

r

P/2

M

Q

tension

Pr/20,55 Pr

P/2P+ -

N

P/2

P/2

P

-

+~1,1P

Quasi-arch

19/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

Comparison

M

Q

N

0,2Pr

+ P

P/2

Pr/20,55Pr

Pr/2Pr

Pr/2

-P/2

P

++

-P/2

P

-+ P

P/2

1,1P

P/2

+-

P/2

P/2+

P/2

+

-

0,7P

P/2

P/2P/2

20/19M.Chrzanowski: Strength of Materials

SM1-05: Statics 4: statically determined bar structures

stop