introduction to stress & bending
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Section 15: Introduction to
Stress and Bending
15-1
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Compressive stress:inner ortion
T
Tensile stress: outer
ortion C
Max stresses near theedges, less near the
ax s
neutral axis y
15-2 From: Noffal
x= by
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at neutral axis andzero at the surface
= (QV)/(I b)
Q= area moment
V= vertical shearforce
h
b
15-3 From: Noffal
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B h vi r f B n n r B n in Bending subjects bone to a combination of
ens on an compress on ens on on oneside of neutral axis, compression on the other
axis)
distance from the neutral axis (see figure)
bending fractures
15-4 From: Brown
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Compressive forceactin off-center fromlong axis
15-5 From: Noffal
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Bending Loading
15-6 From: Brown
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Muscle Activity Changing Stress
s r u on
15-7 From: Brown
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Various T es of Beam Loadin and Su ort Beam - structural member designed to support
loads applied at various points along its length.
Beam can be subjected to concentratedloads or
distributedloads or combination of both.
Beam design
is two-step process:eterm ne s ear ng orces an en ng
moments produced by applied loads
2) select cross-section best suited to resist
shearing forces and bending moments
15-8 From: Rabiei
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6.1 SHEAR AND MOMENTDIAGRAMS In order to design a beam, it is necessary to
determine the maximum shear and moment in the
beam Express V and M as functions of arbitrary position
xalong axis.
These functions can be represented by graphscalled shear and moment diagrams
Engineers need to know the variationof shear and
moment a ong t e eam to now w ere toreinforce it
15-9 From: Wang
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Diagrams
15-10 From: Hornsey
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6.1 SHEAR AND MOMENTDIAGRAMS Shear and bending-moment functions must be
determined for each regionof the beam between
any two discontinuities of loading
15-11 From: Wang
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6.1 SHEAR AND MOMENTDIAGRAMSBeam sign convention
Al h h h i f i n nv n i n i r i r r in
this course, we adopt the one often used byengineers:
15-12 From: Wang
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6.1 SHEAR AND MOMENTDIAGRAMSProcedure for analysis
Support reactions Determine all reactive forces and couple moments
acting on beam
perpendicular and parallel to beams axis
Shear and moment functions Specify separate coordinates x having an origin at
beams left end, and extending to regions of beambetween concentrated forces and/or cou le
moments, or where there is no discontinuity ofdistributed loading
15-13 From: Wang
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6.1 SHEAR AND MOMENTDIAGRAMSProcedure for analysis
Shear and moment functions Section beam perpendicular to its axis at each
distance x - Make sure V and M are shown acting in positive
sense, according to sign convention Sum forces perpendicular to beams axis to get
shear
to get moment
15-14 From: Wang
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6.1 SHEAR AND MOMENTDIAGRAMSProcedure for analysis
Shear and moment diagrams Plot shear diagram (V vs. x) and moment diagram
(M vs. x) ,
above axis, otherwise, negative values are plotted
below axis s conven en o s ow e s ear an momendiagrams directly below the free-body diagram
15-15 From: Wang
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.Draw the shear and moment diagrams for beamshown below.
15-16 From: Wang
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.Support reactions: Shown in free-body diagram.h r n m m n f n i n
Since there is a discontinuity of distributed loadand a concentrated load at beams center tworegions of xmust be considered.
0 x 5 m
+ Fy = 0; ... V= 5.75 N
+ M= 0; ... M= (5.75x1 + 80) kNm
15-17 From: Wang
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.Shear and moment functions 2 ,
+ F = 0 ... V= 15.75 5x kN
+ = 0 ... = 5.75x 2 + 15.75x +92.5 kNm
Check results b a l in w = dV/dx and V = dM/dx.
15-18 From: Wang
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.Shear and moment diagrams
15-19 From: Wang
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6.2 GRAPHICAL METHOD FOR
N TR TIN HEAR AND M MENTDIAGRAMSRegions of concentrated force and moment
15-20 From: Wang
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6.2 GRAPHICAL METHOD FOR
N TR TIN HEAR AND M MENTDIAGRAMSRegions of concentrated force and moment
15-21 From: Wang
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.SOLUTION:
Takin entire beam as a free-bod
calculate reactions atB andD.
Find equivalent internal force-couple
Draw the shear and bendin moment
systems or ree- o es orme y
cutting beam on either side of load
application points.
diagrams for the beam and loading
shown.
Plot results.
15-22 From: Rabiei
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Sample Problem 7.2
Taking entire beam as a free-body, calculate
reactions atB andD.
Find equivalent internal force-couple systems at
sections on either side of load application points.
= :0F 0kN20 = V kN20=V
:02 =M ( )( ) 0m0kN20 1 =+M 01 =M
mkN50kN26
mkN50kN26 33
==
==
V
MV
m ar y,
mkN50kN26
mkN50kN26
66
55
==
==
MV
MV
15-23 From: Rabiei
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Sample Problem 7.2 Plot results.
Note that shear is of constant value
etween concentrate oa s an
bending moment varies linearly.
15-24 From: Rabiei
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