DE THI CUOI HQC KY I NAM HOC 2023-2024
Mon: Ca ky thuat
Ma mon hoc: ENME130620
Be so/Ma de: 01 Be thi co 02 trang.
Thai gian: 90 phut. Ngay thi: 15 /12 /2023
Dirac phep sir dung tai lieu: 01 ta giay A4 viet tay.
TRITCJNG DAI HQC SU PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA DAO TAO CHAT L l/O NG CAO
BO MON CO S 6 THIET KE MAY
Cau 1: (1,5 diem)
Ba lire Fv F2, F3 tac dung len gia treo nhu tren Hinh 1. Biit: Fx - F,F2 = 650 N va F3 = 500 N.
Chon he true toa do Oxy nhu tren hlnh; f va / tuong ung la cac vector dan vi cua true Ox va Oy. Hay:
a. Bieu dien cac vector luc noi tren theo cac vector dan vi T, ]*, F va 0;
b. Neu vector hop luc cua ba luc noi tren co do ldn FR = 600 N va hudng theo true u, tinh gia tri F va
0;
Hinh 1
Cau 2: (1,5 diem)
Cho luc co do ldn F = 80 N tac dung len thanh tai A nhu Hinh 2. Chon he true toa do Oxyz nhu tren
hinh; T ,J va k tuong ung la cac vector dan vj cua true Ox, Oy va Oz. Hay:
a. Bieu dien lire F theo cac vector dan vi 7 ,J, k ;
b. Bieu dien vec ta mo men M0 cua lire F doi vdi diem O theo cac vector dan vi T ,J ,k \
c. Tinh do ldn mo men (M0) cua lire F ddi vdi diem O.
Cau 3: (1,5 diem)
Determine the reactions at the supports A and B for equilibrium of the beam shown in Figure 3.
Cau 4: (1,5 diem)
Cho ca he chju luc, chju lien ket va dang d trang thai can bang nhu tren Hinh 4. Tac dung mot luc
F — 500 N len tay elm. Biet tai A, B, C va B la cac lien ket khdp xoay. Xac dinh lire kep thing dung
tac dung tai E.
dA A
------
-
qi = 800N/m
-6m-
IL
Figure 3
q2 = 600N/m
-4m-
S6 hieu: BM1/QT-PBT-RBTV/02 Lan soat x^t: 02 Ngay hieu lire: 15/12/2023 Trang: 1/2
Cau 5: (2,5 diem)
Cho co cau chuyen dong nhu Hinh 5. Tai thdi diem khao sat nhu tren hinh ve, con truot A dang
trugt v6i van toe va gia toe tuong ung la VA = 8 m /s va aA = 3 m /s2, chot B truot trong ranh. Biet
lien kit gitra thanh AB va con truot A la khdp xoay. Hay xac dinh:
a. Tam van toe tuc thai cua thanh AB; c. Van toe goc thanh AB;
b. Van toe cua diem B; d. Gia tie goc thanh AB.
VA — 8m/s
Hinh 5 Figure 6
Cau 6: (7,5 diemj
The 50 kg wheel is subjected to a force of 50 N. If the wheel starts from rest and rolls without
slipping shown in Figure 6. Determine its angular velocity after it has rotated 10 revolutions. The
radius of gyration of the wheel about its mass center G is kG = 0,3 m.
Ghi chu: Can bo coi thi khong dugc giai thick de thi.
Chuan dau ra cua hoc phan (ve kien thirc) Noi dung kiem tra
[CLO 1.2]: Thu gon dugc mot he lire ve mot tarn thu gon,
phan tich dugc cac thanh phin phan luc lien ket (ke ca luc
ma sat) va thi6t lap dugc cac phuemg trinh can bing cua chat
diem, cua vat ran, he vat ran.
Cau 1,2
[CLO 1.2]: Tinh dugc cac thong so dong hgc (vi tri, van toe,
gia t6c) cua vat rdn, diim thuoc vat ran trong chuyin dong
phang.
Cau 5
[CLO 1.3]: Su dung dugc phuemg phap dong lugng va
phuong phap nang lugng de xac dinh dugc thong so dong
hoc cua vat ran trong chuyen dong phang duoi tac dung cua
lire
Cau 6
[CLO 2.1]: Phan tich va mo hinh hoa dugc mot so ca he
tmh djnh trong ca khi bang cac mo hinh tuong ung.
Cau 4
[CLO 3.1]: Su dung dugc cac thuat ngU tieng Anh ve ca
ky thuat
Cau 3
N g ay - h thang^^n a m 2023 i
Trirang bo mon /C
S6 hieu: BM1/QT-PDT-RBTV/02 Lin sodt xet: 02 Ngay hieu luc: 15/12/2023 Trang: 2/2
TRUONG DAI HOC SU' PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO SO THIET KE MAY
BE THI CUOI KY HQC KY I NAM HQC 2023-2024
Mon: Sue ben vat lieu (Co1 khi)
Ma mon hoc: MEMA230720
Be s6/Ma dS: 01 Be thi co 2 trang.
Thai gian: 90 phut. Ngay thi: 11/12/2023
Dm/c phep su dung 1 t&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.
FiSure 1 Hinh 2
Cau 2: (1,5 diem)
Khdp noi chju luc P = 10 kN nhir tren Hinh 2. Biet chot A dugc lam bang thep co irng suit tiip cho
phep ranow = [r] = 10 MPa. Xac dinh duang kinh d nho nhat cua chot (lam trdn dan vi mm) theo dieu
kien bin ung suit tiip.
Cau 3: (1,5 diem)
Thanh gom hai doan AC va CH co duang kinh tuang ung d\ = 60 mm va di = 40 mm, chieu dai nhu
Hinh 3. Thanh co lien ket c6 dinh tai A, chiu hai lire Pi = 80 kN dat tai C va Pi = 30 kN dat tai H. Biet
thanh lam bang thep co mo dun dan hoi E = 200 GPa. Liy 7t = 3,14.
a. Ve bieu do noi luc va xac dinh tri so ung suat phap phat sinh tren mat cat ngang tai B va D (lam
trdn dan vi MPa)\
b. Xac djnh tri so va hudng chuyen vi cua cac diem BvkD (lay 2 so le, dan 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. Betermine the shear stress at point D (on the web) on the cross section at C (round in MPa)\
b. Betermine the normal stress at point E on the cross section at C (round in MPa)\
c. Betermine the deflection of end A (round to three decimal places, in mm). Given E = 200 GPa.
di Ddi H
7 \ ?
0,5 m _ . 0,5 m }
Pi
0,5 m ( 0,5 m j
Pi Hinh 3
So higu: B M 1/Q T-PBT-RBTV/02 Lan soat xet: 02 Ngay hi?u lye: 1 5 /5 /2 0 2 0 Trang: 1/2
Cau 5: (1,5 diem)
Mot true dac co dirdng ki'nh 80 mm, dugc lam bang thep co mo dun dan h6i trugt G = 75 GPa, chiu
tai nhu Hinh 5. Dau C dugc ngam vao vach cung. Lay n = 3,14.
a. Ve bieu do mo men xoin noi lire cho true va tinh tri s6 ung suit tilp Ion nhit phat sinh trong true
(lam trdn dan vj MPa);
b. Ti'nh goc xoin (tri so va chieu) cua mat cat ngang a dlu A so voi mat cat ngang a dau C (lay 3 sd
le, dan vi do).
Cau 6: (2,0 diem)
True co tilt dien tron (duong kinh d) dugc do tren hai 6 d& tai A va B, dugc lip hai banh rang tai C
va D, chiu tai nhu Hinh 6. Biet true lam bang thep co ung suat phap cho phep aaiiOH,= [<r] = 150 MPa.
Bo qua anh huang cua luc cat.
a. Ve bieu do ngi luc Mx, My va T(con goi la Mz) cho true (idy 3 sd le , dan vi kN.mm);
b. Xac djnh dudng kinh nho nhat cua true (lam trdn dan vi mm) theo Thuyet ben 4 (Thuyet ben thi
nang bien doi hinh dang cue dai).
2 kN.m
Hinh 5 Hinh 6
Ghi chu: Can bo coi thi khong duac gicii thick de thi.
? A \ r
Chuan dau ra cua hoc phan (ve kien thuc) Ngi dung kiem tra
[CLO 1.1]: Nhan biet dugc cac dang chju luc cua ket cau va chi tiet may.
Tinh toan dugc noi luc khi biSt ngoai luc. Ve dugc bilu do noi luc.
Cau 3, 5
[CLO 1.3]: Co kha nang van dung cac cong thuc lien quan de tinh toan ket
cau, chi tiet may nham dam bao dugc do ben.
Cau 2, 6
[CLO 3.1]: Su dung dugc cac thuat ngO tieng Anh dung cho llnh vuc sue
ben vat lieu trong llnh vuc ca khi.
Cau 1,4
NgayT" th an g /2 nam 2023 ,
Tru-ung Bg mon
(ky va ghi ro h<? ten)
So hifu: B M 1/QT-PBT-RDTV/02 Lan soat xet: 02 Ngay h i|u lye: 1 5 /5 /20 2 0 Trang: 2/2
T R U IN G DAI HOC S I/ PHAM KY THUAT
t h An h PHO HO CHI MINH
KHOA DAO TAO CHAT L l/ON G CAO
BO MON CO SO THIET KE MAY
dAp An d e t h i c u o i k y h k i - n h 23-24
Mon: Ca ky thuat
Ma mon hoc: ENM E130620
De so/Ma de: 01 Dap an co 03 trang
Cau 1
F1 — F coscp i + F sincp J
T2 = (- 6 5 0 J t + 6 5 0 ^ ;) N
F^ = ((500 co s4 5 )i — (500 sin 4 5 );) = (25 0a/2 i - 250y[2 j)N
1,5 d
0,25 d
0,25 d
0,25 d
Fr = F1 + F2 + F3 = ( f coscp - 6 50^ + 250V2) 1 + ( f sincf) + 6 5 0 ^
250V T ) ; = ((60 0 cos45 °)t + (600 sin 4 5 °);) N
F coscp — 6 5 0 - + 250V2 = 600 cos45° rp — 527,93 N
F sincp + 650 - — 250V2~ = 600 sin45
fF = 527,93 A
0 ~* ( 0 = 29,23°
0,25 d
0,5 d
Cau 2 1,5 d
a. • A (3; 3; 0) m ; C(4; 0; —2) m -» AC = (1; —3; —2) m
AC = V(1 Y + (~ 3 ) 2 + ( —2) 2 = V l4 = 3,74 m
• F = F. — = 80. — 1 + 8 0 .— / + 80. — fc
j4C 3 ,7 4 3 , 7 4 '' 3,7 4
• F = (21,29i - 64,17; - 42,78/c) A
0,25 d
0,25 d
0,25 d
• OA = (3; 3; 0) m
• 1% = O A xF = ( - 1 2 8 ,3 4 i + 12 8 ,3 4; - 256,38 fc) IV.m 0,5 d
M0 = V (- 1 2 8 ,3 4 )2 + (128,3 4)2 + (- 2 5 6 ,3 8 ) 2 = 314,12 IV.m 0,25 d
Cau 3 1,5 d
• Sa do GPLK
8
X.
a A
4 m 3 m
Qi Q2Q3
r * , i i *' i ' ' 1 ♦ -
y
x
•B
6 m 4 m
• Q1 = 800.6 = 2400 IV; Q2 = 200.4 = 400 IV;
2 2
& = 600.4 = 2400 N
0,5 d
0,25 d
Z fir = 0 Y4 = 00,25 d
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