[email protected] engr-36_lec-19_beams-2.pptx 1 bruce mayer, pe engineering-36: engineering...

[email protected] • ENGR-36_Lec-19_Beams-2.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Chp 7:Beams-2



Beam – What is it? Beam Structural member

designed to support loads applied at various points along its length

Beams can be subjected to CONCENTRATED loads or DISTRIBUTED loads or a COMBINATION of both.

Beam Design is 2-Step Process

1. Determine Axial/Shearing Forces and Bending Moments Produced By Applied Loads

2. Select Structural Cross-section and Material Best Suited To Resist the Applied Forces and Bending-Moments



Beam Loading and Supports

Beams are classified according to the Support Method(s); e.g., Simply-Supported, Cantilever

Reactions at beam supports are Determinate if they involve exactly THREE unknowns. • Otherwise, they are Statically INdeterminate



Shear & Bending-Moment Goal = determine bending

moment and shearing force at any point in a beam subjected to concentrated and distributed loads

Determine reactions at supports by treating whole beam as free-body.

Cut beam at C and draw free-body diagrams for AC and CB exposing V-M System

From equilibrium considerations, determine M & V or M’ & V’.



V & M Sign Conventions

Consider a Conventionally (Gravity) Loaded Simply-Supported Beam with the X-Axis Origin Conventionally Located at the LEFT

Next Consider a Virtual Section Located at C

P

xC

DEFINE this Case as POSITIVE • Shear, V

– The Virtual Member LEFT of the Cut is pushed DOWN by the Right Virtual Member

• Moment, M– The Beam BOWS

UPward



V & M Sign Conventions (2)

Positive Shear• Right Member

Pushes DOWN on Left Member

Positive Bending• Beam Bows UPward

POSITIVE Internal Forces, V & M• Note that at a Virtual

Section the V’s & M’s MUST Balance



V & M Diagrams

With the Signs of V&M Defined we Can now Determine the MAGNITUDE and SENSE for V&M at ANY arbitrary Virtual-Cut Location

PLOTTING V&M vs. x Yields the Stacked Load-Shear-Moment (LVM) Diagram

LOAD Diagram

SHEAR Diagram

MOMENT Diagram

“Kinks” at Load-Application Points



Relations Between Load and V

On Element of Length Δx from C to C’; ΣFy = 0

wx

V

dx

dV

xwVVV

x

0lim

0

Separating the Variables and Integrating from Arbitrary Points C & D

curve LOADunder area

D

C

D

C

x

x

CD

V

V

dxxwVVdV



Relations Between Ld and M

Now on C-C’ take ΣMC’ = 0

Separating the Variables and Integrating from Arbitrary Points C & D

VxwVx

M

dx

dM

xxwxVMMM

xx

21

00limlim

02

curve SHEARunder area D

C

x

x

CD dxxVMM

0



Summary: Load, V, M Relations The 1st Derivative of

V is the Negative of the Load

0

0

xwdx

dV

xx

The Shear is the Negative of the Area under the Ld-Curve

CdxxwV

The 1st Derivative of M is the Shear

0

0

xVdx

dM

xx

The Moment is the Area under the V-Curve

CdxxVM



Recall: Derivative = SLOPE The SLOPE of the

V-Curve is the Negative-VALUE of the Load Curve

The SLOPE of the M-Curve is the Positive-VALUE of the Shear Curve

00

0

xwxmdx

dVV

xx

00

0

xVxmdx

dMM

xx

xydxdy

mdxdy

yx

• Note that w is a POSITIVE scalar; i.e.; it is a Magnitude



Calculus Summary

The SLOPE of the V-curve is the negative MAGNITUDE of the w-Curve

The SLOPE of M-Curve is the VALUE of the V-Curve

The CHANGE in V between Pts a&b is the DEFINATE INTEGRAL between Pts a&b of the w-Curve

The CHANGE in M between Pts a&b is the DEFINATE INTEGRAL between Pts a&b of the V-Curve



Calculus Summary When w is down

• The negative VALUE of the w-Curve is the SLOPE of the V-curve

• The Negative AREA Under w-Curve is the CHANGE in the V-Value

• The VALUE of the V-Curve is the SLOPE of M-Curve

• The AREA under the V-Curve is the CHANGE in the M-Value



Example V&M by Calculus

For the Given Load & Geometry, Draw the shear and bending moment diagrams for the beam AE

Solution Plan• Taking entire beam

as free-body, calculate reactions at Support A and D.

• Between concentrated load application points, dV/dx = −w = 0, and so the SLOPE is ZERO, and Thus Shear is Constant




Solution Plan (cont.)• With UNIFORM

loading between D and E, the shear variation is LINEAR– mV = −1.5 kip/ft

• Between concentrated load application points, dM/dx = mM = V = const.

• The CHANGE IN MOMENT between load application points is equal to AREA UNDER SHEAR CURVE between Load-App points

• With a LINEAR shear variation between D and E, the bending moment diagram is a PARABOLA (i.e., 2nd degree in x).



Example V&M by Calculus Taking entire beam as a

free-body, determine reactions at supports :0AM

0ft 82kips 12

ft 14kips 12ft 6kips 20ft 24

D

kips 26D:0 yF

0kips 12kips 26kips 12kips 20 yA

kips 18yA00 xx AF




The VERTICAL Reactions

Between concentrated load application points, dV/dx = 0, and thus shear is Constant

kips 26Dkips 18yA

With uniform loading between D and E, the shear variation is LINEAR.• SLOPE is constant at −w

(−1.5 kip/ft in this case)




Between concentrated load application points, dM/dx = V = Const. And the change in moment between load application points is equal to AREA under the SHEAR CURVE between points.

ftkip 48140

ftkip 9216

ftkip 108108

DCD

CBC

BAB

MMM

MMM

MMM

ft6 ft8 ft10 ft8




With a Linear Shear variation between D and E, the bending moment diagram is PARABOLIC.

048

ftkip 48

EDE

D

MMM

M

Note that the FREE End of a Cantilever Beam Cannot Support ANY Shear or Bending-Moment

ft6 ft8 ft10 ft8



WhiteBoard Work

Let’s WorkThis Problemw/ Calculus& MATLAB



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix

[email protected] engr-36_lec-19_beams-2.pptx 1 bruce mayer, pe engineering-36: engineering...

Documents

pe engineering

beam supports

beam design

indeterminate slide

virtual member left

right virtual member

arbitrary points c d

structural member