EĞİLME MOMENTİ VE KESME KUVVET DİYAGRAMLARI

dv = - w(x)

dx

kaymanın eğimi = - yayılı yük

herbir noktadaki

kuvvet diyagramı herbir noktadaki yoğunluk

dM = V

dx

herbir noktadaki momentin eğimi = herbir

noktadaki kesme

farklı eğim tarifleri

eğim = sıfır

sabit pozitif eğim

azalan

Positive eğim

artan Positive

slope eğim

sabit negative

eğim

artan Negative eğim

azalan Negative eğim

dv = - w(x)

dx

dM = V

dx

•The slope of the shear force

diagram, at each point, is

equal to the negative of the

intensity of distributed

loading.

•Distributed loading is

positive (w is positive when

it act downward) and

increases from zero to WB.

Thus the slope of the shear

force diagram will be

negative and increases from

0 to –WB.

dv = - w(x)

dx

dM = V

dx

•The slope of the Moment

Diagram at each point is

equal to the shear force (V).

•The slope of the moment

diagram start with a value

of VA then decreases to

zero (tangent become

horizontal) and then

become negative (i.e

shear force is negative) to

a value of –VB.

Regions of concentrated Force and Moment

•When a concentrated force acts downward so the shear force

diagram will jump downward at that particular point.

•When a concentrated moment Mo is applied clockwise, the moment

diagram will jump upward. When Mo acts counterclockwise, the

moment diagram will jump downward.

dM= V dX

∆MAB = MB - MA= VLAB = 4.8 kN (6 m) = 28.8 KN.m

Since MA = 0 thus: MB=28.8 KN.m

MAB is a straight line of slope 4.8

∆MBC = MC - MB= VLBC = -3.2 (2) = -6.4 KN.m

MC = 28.8 – 6.4 = 22.4 KN.m

∆MCD = MD - MC= VLBC= -11.2 (2) = -22.4 KN.m

MD = 22.4-22.4 = 0

•There is no distributed loading (w=0)

•Slope of the shear force at any point (dv /dx)= -

w=0

•Thus the shear force has to be a horizontal strain

line (slope = 0)

•V=p (shear force is positive since it cause a

clockwise rotation, check sign notations)

(dM/dx)=V=P

Thus the moment diagram has a constant slope of value P,

thus it has to be a straight line with a positive slope

At x=0, M= -PL x=L , M=0

örnekler

6

–83. Determine the absolute maximum bending stress in the tubular shaft if di = 160 mm and do = 200 mm .