DE THI CUOI KY HQC KY I NAM HOC 2023-2024
Mon: Sue ben vat li?u (Co khi)
Ma mon hoc: MEMA230720
Be so/Ma de: 01 DS thi co 2 trang.
Thai gian: 90 phut. Ngay thi: 20/12/2023
Dupe phep su dung 1 tit A4 chep tay hai mat.
TRl/ONG DAI HQC SIT PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON CO SOf THIET KE MAY
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 o f the beam at point H.
Cau 2: (1,5 diem)
Cho moi ghep bu long chiu cit nhu tren Hinh 2. Biet bon bu long co cung duong kinh d = 20 mm va
dugc lam bang thep co ung suit tiSp cho phep Taliow = [r] = 15 MPa. Xac djnh gia trj lire P Ion nhat
{lam tron d m vi N) theo dieu kien bin ung suit tiep d6i vdi bu long.
Cau 3: {1,5 diem)
Thanh AD co duong ki'nh d = 50 mm va chieu dai dugc cho nhu tren Hinh 3. Thanh co lien ket c6
djnh tai A, chju hai luc P\= 100 kN dat tai B^kPj- 80 kN dat tai C. Biet thanh lam bang thep co mo
dun dan hoi E = 200 GPa. Liy n = 3,14.
a. Ve bieu d6 noi luc va xac djnh gia trj tuyet d6i ung suit phap lan nhit phat sinh trong thanh {lam
tron dm vi MPa);
b. Xac djnh trj so va hudng chuyen vj cua cac diim B va D {lay 2 so le, d m vi mm).
Cau 4: {1,5 diem)
A box beam is shown in Figure 4. If a bending moment and a shear force acting on the cross section
of the beam are M = 10 kN.m and V= 15 kN, respectively, determine the maximum normal stress and
the maximum shear stress developed in beam {round in MPa).
| EE
S3
P i P i
A B
1 m
CD
1,5 m 1,25 m
Hinh 3 Figure 4
Cau 5: {1,0 diem)
Mot true dac chiu tai xoan M= 20 kN.m, g6m ba doan nhu Hinh 5. Doan AB va CD co cung duong
kinh 60 mm, doan giua BC co duong kinh 80 mm. Bo qua ma sat tai cac goi da. Hay ve biiu d6 noi luc
va xac djnh goc xoan (trj s6 va chiiu) cua banh rang A so vdi banh rang D {idy 3 sd le, dm vi dp). Liy
7t = 3,14. Biet true lam bang thep co mo dun dan hdi trugt G = 75 GPa.
So hiqu: BM1/QT-POT-RDTV/02 Lan soat xet: 02 Ngay hi§u Igc: 15/5/2020 Trang: 1/2
Cau 6: (1,0 diem)
Xac djnh do v5ng tai dau A (lay 2 so le, don vj mm) cua dam chju tai nhu Hinh 6. Bilt E = 200 GPa;
/= 65x 106 mm4.
20 kN
\D 15 kN.m
c.
AB C
1 m 1 m
; !
Hinh 6
Cau 7: (2,0 diem)
True co tiet dien tron (duemg kfnh d) dugc d& tren hai 6 d& tai A va D, dugc lip hai banh rang tai B
va C, chiu tai nhu Hinh 7. Biet true lam bang thep co ung suat phap cho phep (Taiiow= M = 150 MPa.
Bo qua anh hubng cua lire cit.
a. Ve bieu d6 noi luc Mx, My va T (con ggi la Mz) cho true (lay 3 s6 le , don vi kN.mm);
b. Xac dinh duong kinh nho nhit cua true (lam tron don vi mm) theo Thuyit bln 4 (Thuyet ben the
nang bien doi hinh dang cue dai).
y
Hinh 7
Ghi chu: Can bo coi thi khdng rfuoc gicii thick de thi.
Chuan dau ra ciia hgc phan (ve kien thuc) Ngi dung kiem tra
[CLO 1.1]: Nhan bilt dugc cac dang chiu luc cua kit cau va chi tilt may.
Tinh toan dugc noi luc khi bilt 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 tilt may nhim dam bao dugc do ben.
Cau 2, 6
[CLO 3.1]: Sir dung dugc cac thuat ngtr tieng Anh dung cho lmh vuc sue
ben vat lieu trong lmh vuc co khi.
Cau 1, 4
TRUING DAI HOC SPKT TPHCM
KHOA CO KHI CHE TAO MAY
BO MON CO Sti THIET KE MAY
dAp An t h i c u o i k y -h k i - nAm h o c 2023-2024
MON SUG BEN VAT LIEU -NGA Y THI 20-12-2023
MA MON HOC: MEMA230720
De so/Ma de: 01 Dap an co 4 trang
Cau
hoi
Cau 1
(-1,5
diem)
Noi dung
So do giai phong lien ket cho dam:
12 kN
16 kN.m
____
.
r
YakcH v D
1 m_ 1 m . J m v ,1m.
B
Yb
Diem
0,25
Phuong trinh can bang:
£ Mb = 0 -> -Y a x 4 + 12 x 3 - 16 = 0 -A YA = 5 kN 0,25
So do giai phong lien ket cho doan dam AH:
12 kN
z
* A
5 kN
______
*
______
1
k
1 m
C H
1 m
M (con gpi la Mx)
5
V (con goi la Qy)
0,50
Phuong trinh can bang:
Y,Fy = 0 —V + 5 — 12 = 0
I = 0 A M - 5 x 2 + 12x1 = 0
-> V = -7kN
■A M = -2 kN.m
0,25
0,25
Cau 2
C1,5
diem)
So do giai phong lien ket de tim luc cat trong bu-long:
2V
xI I
J—L-
2V
0,25
Phuong trinh can blng:
£ Fx = 0 -> —4V + P = 0 -> V = 0,25P0,50
Dieu kien ben dng suat tiep doi voi bu-long:
V 0,25P ^ r 1 „ r
T = - = < [t] = 15 MPa
A 3,14x202/4 l 1
Giai tim P:
P < 18840 N
Pmax = 18840 N
0,25
0,25
0,25
Cau 3
C1,5
diem)
a. Bieu do noi lire N phat sinh trong thanh:
(a) , kN
0,50
------------
-
---------------
-
----------------
7
-------------------------
7
--------------------------------------------
Gia tri tuyet doi ung suat phap Ion nhat phat sinh trong thanh:
M i Mr. 80xlQ3
3,14x502/4 41 MPa 0,25
b. Chuyen vj doc true cua diem B:
$b — $b/a = ^Lab — Nab-LaB _ (—20X103)X (1X103)
E.A (200x103) x(3,14x502/4)
Diem B djeh chuyen sang trai 0,05 mm
—0,05 mm
Chuyen vi doc true cua di6m D:
Sd = <5d/a = A L ab + h L Bc + A L c o
_ (—2QX103)x(1X103) (80X103)X(1,5X103)
(2 00 x103) x(3,14x502/4) (200X103) x(3,14x502/4)
Diem D dich chuyen sang phai 0,25 mm
Nab-Lab _j_ Nbc-lbc ^ Ncd-Lcd _
E.A E.A E.A ~
+ 0 & 0,25 mm
0,50
0,25
Cau 4
(1,5
diem)
Mo men quan tinh / (hay con goi la Ix) cua tiet dien doi voi true trung hoa x:
, 200X1203 160X803 4
/ =
-----------------------
« 21973333 mm4
, , 12 r 12
Ung suat phap Ion nhat phat sinh tren ti6t dien:
|M| , , , 10X106
= +T-\y\max = +] x 60 » +27 MPa
21973333
Urig suat ti£p Ion nhat phat sinh tren ti6t dien:
*max = ( hogc d im g cong thu c Tmax = )
Trong do:
Q m a x = 200 x 20 x 50 + 2 x 20 x 40 x 20 = 232000 mm3 (Qmax con goi la S™ax)
t = 2 x 20 = 40 mm (t con goi la bc)
Suy ra:
1 5 x l0 3x232000
21973333X40 4 MPa.
0,25
0,50
0,25
0,25
0,25
Cau 5
(1,0
diem)
VS bieu do mo men xoan noi luc T (con la goi Mz)
20 kN.m
\ B C D
^ 0,5 m t 0,2 0,5 m
G
20 kN.m
©kN.m
20
Lieu y: Sinh vien quy woe dau +/- cua mo men xoan T theo chieu nguqc lai van duqc tinh diem
Tinh goc xodn cua mat cat ngang tai A so veri mat cdt ngang tai C:
Va/D ~ ™a/B + ®B/C + 0 C /D — ^A/B + ®B/C ~
-----
^~
---
--------
1
-------
“J“ — —
_ 0 -20X106X0,5X103 -20x1Q6X0,2x103
___
n 9 ? ^ W
75xl03x(l+!)x604 75xl03x(^li)x804 ~ ' ^
®a/d = 0'2g31^18° ~ —12,783° (quay 12,783°cimg chieu kim dong ho)
0,50
0,25
0,25
Ve cac bieu do mo men uon cua dam tirong ung d hai trang thai “jm” va “A”.
20 kN
15 kN.m
Trang thai um”
Cau 6
(1,0
diem)
Trang thai “k”
A
1 m
B C
1 m
15
“ ■v . 35
" -
/cl =1
-------------
;
------
|/c2 =5/3
(m) , kN.m
(M ), m
Tfnh gia tri do v6ng theo cong thuc:
a _ wl\/cl+w2-/c2 _
a A ~ ' El
[(- 1 5 x l0 6)x (2 x l0 3) ] x ( - l x l0 3) + ||x ( - 2 0 x l0 6) x ( l x l 0 3) | x ( - | x l 0 3)
(200x103) x(65x106)
= 3,59 mm (hu&ng xuong ducri)
0,25
0,25
0,25
0,25
So do tfnh:
X A
Ya
y
750
15
100 mm
<
-------
kN.mm 10kN
A C
klN\ 750
200 mm
X D
N ,
200 mm
£>
0,50
= 4kN
= 12 kN 0,25
ve bieu do noi luc van tinh diem
(*), kN.mm 0,25
(m^ , kN.mm 0,25
(t) , kN.mm 0,25
Xac dinh phan lien ket luc tai goi:
Z M x/a = 0 ^ Yd = 6 kN ;
2 MyM = 0 XD = 3 kN
EFy = 0
2 ^ = 0
Cau 7
(2,0
diem)
c//o p/zan xdc phan luc lien ket tai goi.
Bieu do mo men uon Mx, My va mo men xodn T (con goi la Mz):
Z ABc
y
^ z
400
1200^.
~ o "—■
**1200
600
X750 750
w
D
Luu y: Sinh vien quy uac dau +/- cua mo men theo chieu nguoc lai va nhat quan van duqc tinh
diem.
Suy ra: {yjM% + M2 + 0,75T2) = V l200 2 + 6002 + 0,75 x 7502 « 1490,596 kN. mm 0,25