Vijay Gupta

MEC201Mechanics of Solids

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Lecture 18

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Recap: Pre-stressing

(a) Tendon being stresses during casting. Tension in tendon, no stress in concrete.

(b) After casting, the force is released and the structure shrinks.

(c) FBD of tendon. The concrete does not let the tendon shrink as much as it would on its own. This results in residual tension in the tendon.

(d) FBD of concrete. The residual force in the tendon is trying to compress the concrete.

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Recap: A simple example

A concrete beam of cross-sectional area 5 cm×5 cm and length 2 m be cast with a 10 mm dia mild steel rod under a tension of 20 kN. The external tension in steel released after the concrete is set. What is the residual compressive stress in the concrete?

T = 20 kN →σ = 255 Mpa →ε = 1.21×10- 3 →δ = 2.42 mm

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Recap: A simple example

2.42 mm

δc

δs

δs + δc = 2.42 mm

F

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Recap: Failure Under Compression

Buckling

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Torsion

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Shaft carrying torque

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Torsion-bar suspension

Load

Torsion bar

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Torsion-bar suspension

www.rctankhq.com/Project2.htm

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Torsion of a bar

φ zθ r

T

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Torsion of a bar

dz

z

• Plane sections remain plane• Diametrical lines within plane

sections remain straight

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Torsion: Deformation to Strain

A

A1

B1

C

D

O

r

RE1

F1

G1

E

Bγθz = rdφ/dz

τθz

τzθ

γθzdφ

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Torsion: Strain to Stress

γθz = rdφ/dz τθz

τzθγθzA

A1

B1

C

D

O

r

RE1

F1

G1

E

B

dφ

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Torsion: Stress Distribution

τθz = Gγθz = Grdφ/dz

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Torsion: Stress to Load

θ

dθ

dr

Polar (second) moment of area Izz

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Torsion: Stress to Load

τθz = Gγθz = Grdφ/dz For shaft with constant G, R and T:

GIzz is termed as the Torsional rigidity

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Circular Shaft

θ

dθ

dr

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Circular Shaft

θ

dθ

dr

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Hollow Shaft

By geometry: γθz= rdΦ/dzTherefore, τθz = GrdΦ/dz

r varies from R1 to R2, and θ varies from 0 to 2π

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Power shaft

z●+9 Nm

φ = φ1 + φ2

●

−4.5 kNm

●

−4.5 kNm

TMD

φ 0.1 m

φ 0.3 mφ 0.3 m1.5 m 3 m

φ1 φ2

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Power shaft

= 9.82×10-6 m4

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Power shaft

Let us check on the stresses:

Quite safe

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1 m 0.6 m

Φ 10 cm

Φ 6 cm

150 N.m

Φ 5 cm

Φ 2 cm

Another example

F

F

150 N.m

−250 N.m

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1 m 0.6 m

Φ 10 cm

Φ 6 cm

150 N.m

Φ 5 cm

Φ 2 cm

Another example

F

F

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1 m 0.6 m

Φ 10 cm

Φ 6 cm

150 N.m

Φ 5 cm

Φ 2 cm

Another example

F

F

Angle θ2 which represents the counter-clockwise movement of the smaller gear due to gearing alone is 10/6 of θ1 or 0.0085 rad counter-clockwise

Rotation of the right end of second shaft wrt stationary wall is, therefore, 0.0085 rad + 0.12 rad = 0.1285 rad or 7.36 degree.

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A statically indeterminate case

MoM1 M2

Geometric Condition: Φ1 +Φ2 = 0TMD

Equilibrium Condition: - M1 + Mo – M2 = 0

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A Composite Shaft

Shear strain

Shear stress

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Thin-walled shaft

q1 = q2

Shear flow on horizontal surfaces is same as on the vertical surfaces

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Thin-walled shaft

O

h

Relating q to twisting moment T

dT = 2×Grey areaQ = T/2A

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An Example

R 20 mm

R 16 mm

T 100 Nm

This gives τ = (49 N/m)/0.004 m = 12.25 MPa

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Forces and Moments in Beams

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33

Introduction to Beams

• A beam is a horizontal structural member used to support loads

• Beams are used to support the roof and floors in buildings

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x

P

L −x

PV

MV

M

Forces in Beams

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Forces in Beams

z

x

y

Fy

Mz

Mx

FxFz

My

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Forces in Beams

Fx: the axial force that results in elongation (or compression, if it is negative) of the member.

Fy and Fz: shear forces that result in shearing at the section. The shear forces are conventionally assigned the symbol V.

Mz: the axial moment that is the torsion moment that causes twisting of the member. This was the subject matter of the last chapter.

Mx and My:

transverse moments that are termed as the bending moments and cause the member to bend. Moment Mx results in bending the beam in the y-z plane while the moment Mz results in bending the beam in the x-y plane.

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Forces in Beams

Shear forces, VBending Moment, M

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Sign Conmvention

x

z

y

V

M V

M

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Sign Convention

Positive V & M Negative V & M

+_

y

x

+ -

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Sign Convention

Sign of the outward normal to the section

Actual direction of the force or moment

Assigned sign to the shear force or bending moment

Along the + ve coordinate direction

Along the +ve coordinate direction

+ ve

Along the − ve coordinate direction

Along the − ve coordinate direction

+ ve

Along the + ve coordinate direction

Along the − ve coordinate direction − ve

Along the − ve coordinate direction

Along the +ve coordinate direction

− ve

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Sign Convention

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Sign Convention...

Positive Bending MomentNegative Shear force

x

y

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Origin of Shear Force

• Shear Stresses

vary along the height

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Origin of Resisting Moment

Compression near topExtension near bottom

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Origin of Resisting Moment

Net tensile force is zero!

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Origin of Resisting Moment...

Extension near topCompression near bottom

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Bending Stresses in Beams

Net tensile force is zero!

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Beam Supports

Simple Roller Built-in

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Idealized supports

Support type Freedom of motion Reactions presentBuilt-in support No degree of freedom A moment as well as

horizontal and vertical reaction forces

Pinned support

Single degree of freedom - rotation.

Horizontal and vertical reaction forces

Roller support

Two degrees of freedom – rotation and horizontal movement

Only vertical reaction force

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Loads

w N/mP

Concentrated loads Distributed loads

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Loads

Linearly varying distributed load Concentrated moment

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Variations in Shear Force & Bending Moment

y

x

x

M – Px/2 = 0, or M = Px/2

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Shear Force & Bending Moment Diagrams

P

P/2 P/2

V = - P/2

V = + P/2x

x

y

x

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SFD & BMD

SFD- P/2

+ P/2

y

x

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SFD & BMD...

P

P/2 P/2

M – x(P/2) = 0

M – x(P/2) + (x – L/2)P = 0

x

x

y

x

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SFD & BMD...

SFD- P/2

+ P/2

+PL/4BMD

y

x

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SFD & BMD: Another ExampleP P

PP

V = - P

V = 0

P

xP

P

x

y

x

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SFD & BMD: Another ExampleP

PP

P

SFD

- P

+ P

y

x

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SFD & BMD: Another ExampleP

PP

P

M M – xP = 0

M

M – xP + P(x – L/4)= 0

PP

xP

Px

y

x

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SFD & BMD: Another ExampleP

PP

P

SFD

- P

+ P

BMDPL/4

y

x

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SFD & BMD: Another Examplew N/m

wL/2 wL/2

V + wL/2 – wx = 0Or, V = - wL/2 + wx

wL/2V

x

y

x

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SFD & BMD: Another Examplew N/m

wL/2 wL/2

SFD

- wL/2

+ wL/2x

y

x

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SFD & BMD: Another Examplew N/m

wL/2 wL/2

M – (wL/2)x + wx2/2 = 0Or, M = wLx/2 - wx2/2

MwL/2V

xwx

y

x

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SFD & BMD: Another Examplew N/m

wL/2 wL/2

SFD

- wL/2

+ wL/2x

BMDwL2/8

y

x

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Method of Sections:

• Draw an idealized loading diagram of the beam.• Determine the reactions at all supports. If the

reactions cannot be determined, the beam is statically indeterminate and further progress cannot be made without considering the deflections of the beam.

• Determine the number of segments with distinct loading pattern to cover the entire beam. In practice, this means that we segment the beam such that the end of a segment is at the location of a discontinuity in loading pattern.

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• For each of the segments identified above, introduce a cutting plane and draw an FBD of either part of the beam

• Introduce the unknown shear force V and the bending moment M at the cutting plane. These should be shown assuming they are positive according to the sign.

• Determine the expressions for SF and BM by equilibrium considerations, equating the sum of vertical forces and the moment to zero.

Method of Sections:

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SFD & BMD: Another ExampleMoMo/L Mo/L

V = - Mo/L

V = - Mo/L

Mo/L

x

●

MoMo/L

x

●

y

x

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SFD & BMD: Another Example

SFD

- Mo/L

x

MoMo/L Mo/L●y

x

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SFD & BMD: Another ExampleMoMo/L Mo/L

M - Mox /L = 0

M + Mo – Mox/L = 0; or M = Mox/L – Mo

M

M

Mo/LMo/L

x

●

MoMo/L Mo/Lx

●

y

x

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SFD & BMD: Another Example

SFD

- Mo/L

x

MoMo/L Mo/L

BMDx

+Mo/2

-Mo/2

●y

x

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SFD & BMD: Another ExamplePP

PL

V = - Px

PPL

y

x

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SFD & BMD: Another ExamplePP

PL

xV

- P

y

x

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SFD & BMD: Another ExamplePP

PL

x

MP

PL M + PL – Px = 0,M = - P(L – x)

PM + P(L – x) = 0,M = - P(L – x)

y

x

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SFD & BMD: Another ExamplePP

PL

xSFD

- Px

BMD

- PL

y

x

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SFD & BMD: Another ExamplePP

PL/2

PL/2 V V = 0

V V = - PPL/2P

y

x

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SFD & BMD: Another ExamplePP

PL/2

xSFD

- P

y

x

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SFD & BMD: Another ExamplePP

PL/2

PL/2M

PM

M + PL/ 2 = 0

PL/2P

xM + PL/ 2 – P(x – L/2) = 0, orM = - PL/ 2 + P(x – L/2)

y

x

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SFD & BMD: Another ExamplePP

PL/2

xSFD

- PxBMD

- PL/2

y

x

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wLV

x

wL2/2

SFD & BMD: Another Examplew N/m

wL

wx

V + wL – wx = 0, orV = - wL + wx

wL2/2

wL

y

x

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SFD & BMD: Another Examplew N/m

wLxSFD

- wL

wL2/2y

x

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SFD & BMD: Another Examplew N/m

wLM

M + wL2/2 –wLx + wx2/2 = 0or, M = wLx - w(L2 + x2)/2

wL2/2

wL

wL- wxx

wx

wL2/2

y

x

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SFD & BMD: Another Examplew N/m

wLxSFD

- wL

xBMD

- wL2/2

y

xwL2/2

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Comparison: Simply Supported vs. Cantilevered

w N/m

- wL/2

- wL2/2

w N/m

- wL/2

+ wL/2

wL2/8

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Area Method

R1 R2

●

M0

P

AB DC

q(x) N/m

E

x

M

Vdx

M+dM

V + dV

SF as well as BM does not change along a beam segments which does not carry any load

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Area Method: At the location of concentrated force

R1 R2

●

M0

P

AB DC

q(x) N/m

E

dx

M

xV

V + dV

M+dM

P

At the location of a concentrated load, the shear force jumps up by a value equal to the load (acting downwards), but there is no change in the bending moment.

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Area Method: At the location of concentrated moment

dx

M

V

●

x

M0

M+dM

V + dV

R1 R2

●

M0

P

AB DC

q(x) N/m

E

At the location of a concentrated moment, the bending moment jumps down by a value equal to the applied moment, but there is no change in the value of the shear force across such an element.

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Area Method: At the location of a distributed load

q(x)

M+dM

x

V + dVM

Vdx

R1 R2

●

M0

P

AB DC

q(x) N/m

E

The rate of change of shear force at a location is equal to the (downward) load density at that point. Similarly, the rate of change of bending moment is equal to the (negative of) shear force at that point.

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Area Method: Drawing SFD• First draw the loading diagram of the beam.• Calculate the reaction at the supports.• Start drawing the SFD at a point slightly left of the

left support where the shear force is taken as zero.• Travel along the beam to the right, modifying the SF

using the rules given below.• Shear force does not change over the segment of the

beam with no external load.• The presence of a concentrated moment does not

change the value of shear force.

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Area Method: Drawing SFD• A concentrated force causes a jump in SFD at the

location of the force: go up for every load downwards, down for every load upwards.

• The slope of the SFD at any location is equal to the distributed load density (load per unit length). Positive slopes for loads downward, and negative slopes for loads acting upwards.

• Change in SF between any two locations is equal to the area under the distributed load curve. Positive changes for negative areas and negative changes for positive areas.

• Proceed till you are slightly right of the right support where the SF should be zero again. This last is a check on calculations.

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Verifying Area Method: SFD

- P/2+ P/2

P/2 P/2

P

PP

PP

-PP

wwL/2 wL/2

- wL/2

wL/2

- Mo/L

Mo

Mo/L

Mo/L●

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PPPL

xV

-PPPPL/2

x-P

x

-wL

w

wL

wL2/2

Verifying Area Method: SFD

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Constructing SFD100 N 100 N 100 N

200 N/m

150 150

x

-150

+50+100

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• First complete the shear force diagram of the beam. • Start at a point slightly left of the left support where

the bending moment is taken as zero.• Travel along the beam to the right, modifying the BM

using the rules given below.• The slope of the BMD at any location is equal to

(negative of) the SF value at that location• Change in BM between any two locations is equal to

the (negative of) the area under the SF curve.

Area Method: Drawing BMD

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• Concentrated moments cause a jump in BMD at the location of the moment.

• Go up for every negative concentrated moment, down for every positive concentrated moment.

• Proceed till you are slightly right of the right support where the BM should again be zero again. This last is a check on calculations.

Area Method: Drawing BMD

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- P/2

+ P/2

+PL/4

P

P

P

P

- P

+ P

PL/4

Verifying Area Method: BMD

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Verifying Area Method: BMDw N/mwL/2 wL/2

- wL/2

+ wL/2

wL2/8

- Mo/L

+Mo/2

-Mo/2

MoMo/L Mo/L

●

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PP

PL

- P

- PL

PP

PL/2

- P

- PL/2

Verifying Area Method: BMD

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Constructing BMD100 N 100 N 100 N

200 N/m

150 150

-150

+50+100

+150

+150

206.25200

50

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Another Example

20 kN

− 50 kN

− 10 kN

30 kN

40 kNR1

20 kN.m

●

20 kN/mR2

40 kN

0.5 m

x, m1 2 43

A D EF

CB

− 10 kNm

40 kNm45 kNm

25 kNm

30 kNm

x

V

x

M