TRUING DAI HOC SU' PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO Sd THIET KE MAY
DE THI CUOI KY HQC KY I NAM HOC 2023-2024
Mon: Sue ben vat lieu (Co khi)
Ma mon hoc: MEMA230720
Be so/Ma de: 01 B6 thi co 2 trang.
Thai gian: 90 phut. Ngay thi: 11/12/2023
Dir pc phep sir dung 1 tot A4 chep tay hai mat.
Cau 1: (1,5 diem)
A beam is subjected to loads as shown in Figure 1. Determine shear force and bending moment on
the cross section of the beam at point D.
Fi§ure 1 Hinh 2
Cau 2: (1,5 diem)
Khbp noi chiu lire P = 10 kN nhir tren Hinh 2. Biet chot A dugc lam bang thep co ling suat tiep cho
phep Tatiow = [t] = 10 MPa. Xac djnh duang kinh d nho nMt cua ch6t (lam trdn dem vi mm) theo di£u
kien bln trng suit tiep.
Cau 3: (1,5 diem)
Thanh gom hai doan AC va CH co dtrcmg kinh tuang ung d\ = 60 mm va dz = 40 mm, chilu dai nhu
Hinh 3. Thanh co lien ket co dinh tai A, chiu hai lire P\ = 80 kN dat tai C va Pi = 30 kN dat tai H. Bi6t
thanh lam bang thep co mo dun dan hbi E = 200 GPa. Ldy 7t = 3,14.
a. Ve bieu do noi luc va xac djnh trj so tog suat phap phat sinh tren mat cat ngang tai B va D (lam
trdn dem vi MPa);
b. Xac djnh trj s6 va hudng chuyen vj cua cac diem BvaD (idy 2 so le, dem vi mm).
Cau 4: (2,0 diem)
A beam with cross section of I-shape is supported and subjected to loads in Figure 4.
a. Determine the shear stress at point D (on the web) on the cross section at C (round in MPa);
b. Determine the normal stress at point E on the cross section at C (round in MPa);
c. Determine the deflection of end A (round to three decimal places, in mm). Given E = 200 GPa.
0 5 m 0 , 5 m
P i
° ’ 5 m j ( 0 ,5 m
Hinh 3
Figure 4
So higu: BM1/Q.T-PDT-RDTV/02 Lan soat xet: 02 Ngay hifu liyc: 15/5/2020 Trang: 1/2
Cau 5: (1,5 diem)
Mot true ddc co dudng kinh 80 mm, dugc lam bang thep co mo dun dan h6i trugt G = 75 GPa, chiu
tai nhu Hinh 5. Bau C dugc ngam vao vach cung. Lay jt = 3,14.
a. Ve bieu do mo men xo&n npi luc cho true va ti'nh tri so ung suat tiep ldn nhat phat sinh trong true
{lam tron dem vi MPa);
b. Tinh goc xoan (tri so va chieu) cua mat cat ngang a dlu A so vdi mat cat ngang o diu C {lay 3 so
le, dan v; do).
Cau 6: {2,0 diem)
True co tiet dien tron (duong kinh d) dugc do tren hai 6 d5 tai A va B, dugc lap hai banh rang tai C
va D, chiu tai nhu Hinh 6. Biet true lam bang thep co ung suit phap cho phep oanow- [a] = 150 MPa.
Bo qua anh hucmg cua luc cat.
a. Ve biiu d6 noi luc Mx, My va T (con goi la Mz) cho true {lay 3 sole, dom vi kN.mm);
b. Xac djnh duong kinh nho nhat cua true {lam tron dan vj mm) theo Thuyet ben 4 {Thuyet ben the
nang bien doi hinh dang cue dai).
Hinh 5 Hinh 6
Ghi chu: C an bo co i thi khong du a c g id i thick de thi.
Chuan dau ra cua hoc phan (ve kien thuc) Noi dung kiem tra
[G 1.1]: Nhan biit dugc cac dang chju luc cua ket cau va chi tiet may. Tinh
toan dugc noi lire khi biit ngoai luc. Ve dugc bieu do noi luc.
Cau 3, 5
[G1.3]: Co kha nang van dung cac cong thuc lien quan de tinh toan ket cau,
chi tilt may nhim dam bao dugc do ben.
Cau 2, 6
[G3.1]: Sir dung dugc cac thuat ngu tilng Anh dung cho linh vuc sue ben
vat lieu trong linh vuc co khi.
Cau 1,4
N gay f th an g ^ Z jiam 2023
Truong Bg mon
So hifu: BM1/QT-POT-ROTV/O2 Lan soat xet: 02 Ngay higu Igc: 15/5/2020 Trang: 2/2
TRl/CiNG DAI HOC SPKT TPHCM
KHOA CO KHI CHE TAO M AY
BO MON CO S 6 THIET KE MAY
dAp An t h i c u o i k y - h k i - nAm h o c 2023-2024
MON s u e BEN VAT LIEU - NG AY THI 11-12-2023
MA MON HOC: MEMA230720
D es6 /M ad e:0 1 Dap an co 4 trang
Cau
hoi
Noi dung Diem
Sa do giai phong lien ket cho dam:
30 kN/ra 30 kN 0,25
Cau 1
C1,5
diem)
Phucmg trinh can bang:
£ Mr = 0 -» - Y a x 4 - 30 x 2 + 30 x 4 x 2 = 0 -> YA = 45 kN 0,25
So do giai phong lien ket cho doan dam AD:
30 kN/mr
45 kN
(con goi la Mx)
V (con goi la Qy) 0,50
Phuong trinh can bang:
£ Fy = 0 - V + 45 - 30 x 2 = 0
£ MD = 0 ^ M - 45 X 2 + 30 x2x1 = 0
V = - 1 5 kN
-> M = 30 k N .m
0,25
0,25
Cau 2
(7,5
diem)
Sa do giai phong lien ket de tim lire cat trong chot A:
. V
i i
I l
l I P= 10 kN
-----
►0,25
Phuong trinh can bang:
£ Fx = 0 -> - 2 V + 10 = 0->P = 5kN 0,50
Dieu kien ben ung suat tiep doi voi chot A:
t = - = < [t] = 10 MPa
A 3,14xd2/4 L J
Giai tim d\
d > 25,24 m m
Chon: d = 26 mm
0,25
0,25
0,25
Cau 3
(7,5
diem)
a. Bieu do noi lire V p h at sinh trong thanh:
0,5 m r 0,5 m
G
i/
80 kN
. 0,5 m 0,5 m
30
©
30 kN
N) , kN
50
U"ng sudt phap phat sinh tren mat cat ngang tai B:
, = 3,14x602/4 * “ 18 M P a SUClt m n )
U‘ng sudt phap phat sinh tren mat cat ngang tai D:
0,50
0,25
„ _ n d _
°d — ~r —
30x10s
Ap 3,14x402/4 24 MPa (U’ng suat keo)
Diem D dich chuyen sang trai 0,03 mm
0,25
0,25
b. Chuyen vj doc true cua diem B:
o o a, Na b .Lab (—50 x10s) x(0,5x103)
SB — &b/a — ALa b — — 7---------77
-----------
77— ^ 0,04 m m
B B/A AB E.AAB (200x103) x(3,14x602/4)
D ilm B djeh chuyln sang trai 0,04 mm
C huyln vi doc true cua d ilm D:
SD = SD/A = ALac + a l cd = + t o g * =
(- 5 0 x l0 3) x ( lx l0 3) , (30X103)x(0,5x103) _
— —
---------
-
--------------
1—
--------
-
—
--------
-
— k, —(j (j4 mm
(200x103) x(3,14X602/4) (200x103)x(3,14x402/4)
0,25
Cau 4
(2 ,0
diem)
So do giai phong lien ket cho doan dam AC:
--------
- ! c
A
y
1 m B C \
0,5 in
M (con goi la Mx)
V (con goi la Qy)
Phuong trinh can bang:
Y.Fy = 0 ^ - V - 2 0 = 0 -> F = —20 kN
£ M c = 0-»M + 15+ 20x1,5 = 0 -> M = - 45kN.m
a. U’ng su it tilp tai diem D thuoc phln tilt dien doc tren mat cat ngang tai C:
t d = (hoac dung cong thirc t d = )
I.t lx'Dr
Mo men quan tinh / (hay con goi la Ix) cua tiet dien doi voi true trung hoa x:
, 200X3403 2X90X3003 nrnnrrrr*, 4
/ =
----------------------------
= 25006 666 7 m m 4
12 12
Trong do:
Qd = 200 x 20 x 160 = 64000 0 m m 3 (QD con goi la S%)
t = 20 m m (t con goi la bc)
Suy ra:
20x103x640000 „ .
t d = —
..........
.
» 3 MPa
250066667X20
b. U’ng suat phap tai diem E tren mat cat ngang tai C:
°E ~T-\y^\ = ~ ^ o e e L X 170 * ~ 31MPa (.U’n g Sudtnen)
(SVcd the dung cong thuc: oE =
------
—. \yE\ van duac tinh diem)
A A ' *
c. Do vong tai dau A cua dam:
Ve cac b ilu do mo men u6n cua dam tuong ung o hai trang thai “m” va
0,25
0,25
0,25
0,25
0,50
20 kN
Trang thai “my
Trang thai
. :
Pv=\
1 m
B |C
............
F
0.5 m 0.5 m
‘“l 20_
\ © r
(02
/ c i ^ / 3
a*1
7 II
/
______
/c2=5/3
M), kN.m
(M ), m
0,25
Ti'nh gia tr| do vong Aa theo cong thuc:
A c°l fcl+c°2-fc2+c>J3-fc3 _
^A~ El
f(Dx(_20xlo6)x(lxlo3)lx(- f xlo3)+[(_35xlo6)x(lxlo3)]x(_I xlo3) +[(i)x( - 20xlo6) x (l><lo3)lx(-f><103)
(200xl03)x (2 50066667)
= 1,516 m m (hu&ngxuong dw&i)0,25
Cau 5
(1,5
diem)
a. Ve bieu do mo men xoan noi lire T (con la goi ML)
6 kN.m *
Y2 kN.m
Y
c y B V
0,4 m 0,6 m
4
©
~ T 3©kN.m
Licit y: Sinh vien quy uac dau +/- cua mo men xoan T theo chieu nguqc lai van duqc tinh diem
Urig su&t tiep ldn nhat phat sinh trong true:
max ~ , Pmax = t t t zn
----
r X 40 « 40 MPa
; — ( l > o «
(Sink vien co the sir dung cong thuc: Tmax \T\r 4X106
0,2d3 0,2x80s 40 MPa)
0,50
0,50
b. Ti'nh goc xoan cua mat cat ngang tai A so vdi mat cat ngang tai C:
M _ T(AB).Lab+T(BC).Lbc _ —2X106X0,6X103+4X106X0,4X103
Wa/C -
--------------
----------------------
GJ 7 5 xl0 3x (^ ll)x 80 4 0,0013 rad
0a/c — 0'00^ * 18° ~ 0,076° (quay nguqc chieu him dong ho)
3,14
0,25
0,25
So do ti'nh:
0,50
l F y = o
ZFx = 0
-» Ya = 5 kN
-> XA = 8 kN
Cau 6
(2,0
diem)
Xac dinh phan lien ket luc tai goi:
Y.Mx/A = 0 ^ Yb = 10 kN
!M y/A = 0 - > = 2 kN
Luu v: Sinh vien co the tinh cac gia tri mo men noi luc tai C, D roi ve bieu do noi luc van tinh
diem cho phan xac dinh phan luc lien ket tai goi.
Bieu do mo men uSn Mx, My va mo men xoan T (con goi la Mz):
y
0,25
('Mx), kN.mm 0,25
(m^ , kN.mm 0,25
Q ) , kN.mm 0,25