Mechanics of Deformable Bodies
The wooden bar consists of two segments, each of length L. One
segment has a square cross section of width d, the cross section of the
other segment is a circle of diameter d. The working stress for the wood
is tw=5M Paand the shear modulus isG=0.5GPa. Using L=0.6m and
d=50mm, determine (a)the largest torque T that can be safely applied;
and (b)the corresponding angle of twist for the bar. = s
| ne 4 xlb
114s 3S iY
= 3600 fries) = | 702 Jan
3140
3
FA Spaing ®, Trax = KX150-299 Xi, 4 x10-] 08s)
3.14 x¢to)3 Yxlo-y (o
11448 oe e
— 1203).)2 (10) =| 5/428 4 porn
3)4d
Fal Spuingc, Trnava
‘|
Jie x 195-120 x20, te xF5-I 0-615
= ec er
UXFt5-Y
2.14 x(20)3
toh
= 312/92 ( |. 19938) mS
25120 gas
= [14381 oon” EXAMPLE:
15. Previde a Load and Moment Diagram
for the following Sd) Given dota
44
yor
Vth) AHS
—x¢ee)
—|380 = the load diagyam
|. The Sheon diagram tO RB" ic,
Urtheodnly pura, thyy tthe, load Is A
wn} Poserlay disttivuted uptiand abe 975 inlet , + 18666
Moagnikuce Oc
NO
3 ee = 225 Ibletxload in
Segment Bc’.
2. A douanuiond point eovce ach at |
Point *c” With raognitude oF 900 Ib
No lead fn Segment CD"
3. Anorhes GonCenktalad Fosce ic acki Leah ,
down ord ofp’ witha magnitude | Sess! uogisen
OF Qo0 th. 1s 14 :
4. the load Sa’pe’ fs uni foamy distiiiutd —aegolb-Me |
downwand of o Magnitude oF ig00 lof
5 ea) ~ 120 bet
5. AN upwond tood ts contenttolad at’ | * “960 o--
i!
j Het 2a 4et yer uet |
\
|
ane ac"
“EB sits magoitude of Momerk dioggam
> Y80+41380 = 1866 lb
é 8 - ~ 120 Wek fs digtyiowled UNiFoonLy
Oven the $0 "ee" the moment diagéarn
I. Na =0-
2 Ng= Mat Asean sheon diagsan.
SD Mas 0 + + (900) = 1Bo0 Ib-b.
Mes Mat Aveo tq sheon diagsam.
7) Ne= Lov +: 400(2)= 3600 to-BE.
4 Mp= Mc+ Aven ta Shean diagyan.
> Mp= 2600 +6 = 3600 Ib-eE.
5. Ne> Mp Agee io Shean diag jaw
) Ne = 3600 + 4 (90041380) Cup] > ~ 960 Ib-EL.
6. Nps Het Areain Sheor diagjar
=) Ne = — 960 +> @e(Use) (4) = 0-
4. The Shage oF moment diagian fn AB’ Ps Lpusord
porobola, uiitty Vantin ak A, uchile Mazon 408" and
Wat gontal 49°Co' Pos Segment “DE’ tte ciagiam fs dovanusoxd
Poxcbolo. with Venton ot GG Fo the Polat whue. the Orkendad
Sheos fn Ok” Fokerech, the fine of Zes0 Sheoy.
g- The moment dia.gsoum intr isa downs Ponoko
wots Venter ob “p’ A flat steel bar, | inch wide by '4 inch
thick and 40 inches long, is bent by
couples applied at the ends so that
the midpoint deflection is 1.0 inch.
Compute the stress in the bar and
the magnitude of the couples. Use E =
29 x 106 psi. (Practice at Home)
=
EZ
‘Cross Section Froncy Pyiha gor ug theo wm | —
4" % 20 + (ay
ye 20+ ee 24)
> 3 200 -S ins Now ,
a
in
LE =
— R
2 amie woe aR)
OO —=
oe Pe [188.3 Ab~in |
Ged C=
aw
5 .
oe)
= - (Gee. Es 2 [3-08 10” psi
(ay
[18-08 KESi |
ry 6. A50-mm diameter bar is used as a
simply supported beam 3 m long.
Determine the largest uniformly
distributed load that can be applied
over the right two-thirds of the beam
if the flexural stress is limited to 50
MPa. (Assignment)
w N/m
3m
Ry = (2/3) w R. = (4/3) w EXAMPLE:
8.
The right-angled frame shown in the
figure carries a uniformly distributed
loading equivalent to 200 N for each
horizontal projected meter of the
frame; that is, the total load is 1000 N.
‘Compute the maximum flexural stress
at section a-a and section b-b if the
cross-section is 50 mm square. Taksroy epalhbytum of forces
Ry + Rg = 1900
Raxo + Rex 6 —1000% 2 _ oO
a
Rp = tooo
& “gq 7 Seo N
Rp = (00 0 -500- 500N
SS constdes Seduon a-a,
consider two atght Aoiandes
Cse= XL 4
ae
ne HAP km
? =
Consitles Moment,
M- XR, - 200% % x
2
= 24 xX500 - QOOXRAXLH _ Cau
2 =
Mootinu ms Heresal deieen ak a,
4. Me © 2x 1006 x 50 |
77 ss ? 7
Sox 593 = A445 Mpy
Wa =
&
ee 5 haw
Me 4XRg 80 K YxY
Q
~ ~ : te 456 N.
M- LAxsoo Q00x &x 90 : _ i) } ee
Manian flexwal Sus at bbs fi, Me
-
+ A5Ex (000% 50
2
0x 503
IR
= Mb Bey MPa
= EXAMPLE:
3.
Nails having a total shear strength of
40lb are used in a beam that can be
constructed either as in Case | or as
in Case Il (shown in the figure). If the
nails are spaced at 9in, determine the
largest vertical shear that can be
supported in each case so that the
fastener will not fail. (Practice at
Home) To determine:
The largest value of vertical shear that can be supported in each case.
Given Data:
Total shear strength=40 |b
Nail are spaced at 9in Calculation:
Calculate the value of moment of Inertia about the neutral axis.
As the cross-section are same for both the cases.
Case 1: by =3in, hy =S5in
Case 2: bo = 1 in, hy = 4 in
On substituting values on the formula for inerti;
Moment of inertia (I) = (8) —2( 44),
Moment of inertia (I) = (4 x3x ¥)-2[4 x 1x47]
Moment of inertia (1) = 20.58 int Case 1:
In this case only the top flange above the neutral axis or the bottom flange below the neutral axis
considered to go under shear as each of the flange is being attached to web through nail.
So for one of the flanges:
Q=Ay
where
a = Area of the top flange or bottom flange which is in shear due to nail
‘y = The dis tan ce of center of flange consider from neutral axis
QO = yA =2.25 x (3x 0.5) = 3.375i0
The value of maximum shear can be calculated by
vo
T
4 _ ¥G375)
9 20.58
V = 27.1016 In this case the cross section of each flange (1inx0.5in) is attached to web at fourends. Hence any one of
flange can be considered for maximum shear.
Thus,
Q=Ay
where
a= Area of the flange
“y = The dis tan ce of center of flange consider from neutral axis
Q=yA =2.25 x (1 x0.5) = 1. 125in?
The value of maximum shear can be calculated by
vo
qT
sam _ V (1125 in?)
Vin 2058 int
V = 81.3016
The final value calculated as
For case 1:
The largest value of shear is 27.10 Ib
For case 2:
The largest value of shear is 81.30 Ib The wood beam has an allowable
shear stress of Tallow = 7 MPa.
Determine the maximum shear force
V that can be applied to the cross
section. (Practice at Home)
50mm 50mm
+400 mm—-+—+ aoomm
son
Figure
Calculate Moment of
merta x wooden beam
Using Relation
j- BP. t#
12 fe.
J z= [20x 2003 raat foe?
Ap. ya —=
GF = (25 8 po* pas ¥
O= Ag, + 2(A¥ ) =f Go x joo)x 75 ¢ af (ree x50) 50 |
Q = 8h So oo mh &
“Maximum Shear feorce (Vv)
ze =— V x 8} S5c0cCO (i
NE Goh all ablowebhLe Shaay
lo“x (50450) Stoess (6) = 3 mPa
Y= Feo-.6 KN | 60 mm] To A’nd.
Ten Kala s- Y Normal stress +4
J tS Nowal stiees «4 B
Fz 20bN
bb - 60am
he lgom~
Balutseary Sa
New secon moduli utlct be
S= blr _ 60X1f 92
6 Se
S™ 3.24 xP m3
ler BOkN
Ny RED = 4 3
$0. Ofret normal shee (30% He) 008 2220
60X 1g ae
Need Glewlak& Bending Moment.
=S8 X50 - 24x 760
EH = 29X60
Naw “ay SAth = 8% SY x/o3 feeiege
“ So Fibs
Bending Shep =
= /& S$ me,
Of Neral yee, at /2
sme ~ “ae 4 Bending Skt
IW) @e0 Regul te
Rome Shvhns 22222 + 16662
Thuy [Normal shetala — joeee me) gw larly,
TO by
Normal stem apg wlll siltin
= Wire name ;
bf re) ~ shard
2:92 ~ \GoF
—_
{qe — 19 (G48 boke | The cross section of the machine part
is a square, 5 mm ona side. If the
maximum stress at section m-n is
limited to 150 MPa, determine the
largest allowable value of the
eccentricity e. Given
Area of square cross section A = 5? = 25mm? = 25 x 10m?
Maximum allowable stress at section M-n Gna: = 150 MPa
Force acting on machine part P = 250 N
To Find
the largest allowable value of eccentricity. Calculate the sectional modulus (Z)
3)
z= 2 = SAY 90.833 x 10-%m
Allowable stress can be calculated
—MyP
Cmax = 3 +5
where
M = moment due to eccentricity
P= force
A= cross sectional area
Z = section modulus
puiting all value
ona = BE +E
= —250me 250
150. 10" = Boao? * Fx
e = 11.66 x 109m
e = 11.66mm
The maximum allowable eccentricity is 11.66mm