Homework Set #2: Internal Loadings and Stress
Due at 3:55pm on Thursday, February 19
Be sure to include a FBD for each problem. Do NOT just annotate
the figure.
1. Determine the resultant internal normal and shear force in the member on (a) section a-a
and (b) section b-b. Both sections pass through the centroid at A. The 3 kN load is applied
along the centroidal axis of the member.
2. Determine the resultant internal loadings at points C and D. Assume the supports at A and B
are vertical.
1)
∞
y
a
후
t
3 KN
τ
무
.
[Ex
=
Na - 3kN = ) Na = 3 kN
0:
E Fy = 0 :
"
Va
=
0
φ
.
∞
£
3kN
150o'ou
p00
↓
on
#Mani
80kN
*
[Fy
{MA
DN
Fini
Ay
(Ay + By
120
-
=
1. 5 kN
80)kN
200kN
0 : 120 kN
=
-
Vy-3bNsin (30]
0 :
2 m
·
-
80kN
+ 480)kNm
By (20
·
6m + By 8 m
=
i
In
=
=
. 59kN
2
=
VB
& Fy = 0 :
Ay + By
em
Ay
b120
Nb
Nb
d 2 okw
)
Nb-3kN(os30 = 0
n
2
2
[ Fx = 0 :
↑
8
By
=
90 KN
Ay + 90kN
ume M
Ay
EFy
=0 :
=
=
200kN
IOKN
1l0kN-120bN + Vp
VB = 10 KN
EF+= O : ND = O
& Mp = 0 : Mp + (120 4
·
13 33 kW
.
freedoon Pack
MD = 180 kN m
Next
Ʃ F+=O : Nc = 0
EFy
=0
:
Vit 90 KN いふ 33 kN
-
80 kN
Vc = 3. 33kN
5) Mc =
-
Mc
-
13 33 23
.
.
-
80 2 + 90 7
.
Mc = 191 . 1l kN m
.
.
-
110 6) kN-m
·
MEEG 215-010/080
Davis
Spring 2026
2
3. A 155-lb person stands on a hard floor wearing fancy, pink high-heeled shoes. If the heel
has the dimensions shown, determine the average normal stress they exert on the floor
and compare it with the average normal stress developed when the same person is
wearing sensible, flat-heeled shoes. Assume the entire weight is supported only by the
heel of one shoe.
4. Determine the largest intensity w of the uniform loading that can be applied to the frame
without causing either the average normal stress or the average shear stress at section
b-b to exceed σ=30 MPa and τ = 32 MPa, respectively. Member CB has a square cross
section of 40 mm on each side.
Mechanics of Solids
二
)
flat
heel
shoe
r = 0
.
3
ra 12 in
A
=
+
.
O
j : : psi
芝 しで + 0 5 2 φ
こて
.
26 + 12 = 3
、
Gelay
2㎡
=
.
H - C 2 inc
46
.
44 .7psi
”
… 17
.2
=
몸
y)
A : (0 0
.
-
-b
,
→
6m
∞
-
し
( os θ= 0 = 0 . 8
;
6
B = (C 6
C
( 8, 0
82
LiB = 5
=
-
+
Sin θ= V0 = 0 6
.
R= 6 w
= 1am
ㅐ
。
EMA
w6
0 :
=
)
( 3 ) = 18
w
T(0 8)
.
6.+ (0 8) : 4 8T
.
.
4 85 = 58 w
.
T こ 3 25
w
,
A- YOomay
.
YOuon = 16 COmm ? =
T
σ =π こる
1
-1
-
6
θ= 2343 75 w
,
0 = 30MPa
=
30
てる 4 る . 75 = る G
W = 1280
=
+ 1
106
ab
NVm
12 8 KNrm
.
.
1
. 60103 ㎡
MEEG 215-010/080
Davis
Spring 2026
3
5. Section 080 Only: Determine the resultant internal loadings on the cross section through
points C and D of the pliers. There is a pin at A and the jaws at B are smooth.
Mechanics of Solids