Question 1

simple beam AB with an overhang BC is loaded as shown in the figure. The bending moment at the midspan of AB is approximately:

(A) 8 kN ∙ m
(B) 12 kN ∙ m
(C) 17 kN ∙ m
(D) 21 kN ∙ m

B

Accepted Answer

B

Question 2

cantilever beam is loaded as shown in the figure. The bending moment at 0.5 m from the support is approximately:

(A) 12.7 kN ∙ m
(B) 14.2 kN ∙ m
(C) 16.1 kN ∙ m
(D) 18.5 kN ∙ m

D

Accepted Answer

D

Question 3

The simply supported beam ABC shown below is acted upon by applied axial force P at B; by linearly varying distributed load q on BC; and by applied moments

. Reaction forces are given in the figure
In terms of load variable P. The axial force (N), shear (V), and bending moment (M) at mid-span are (in terms
Of variables L and P):

Stresses and strain

D

Accepted Answer

D

Question 4

T-shaped simple beam has a cable with force P anchored at B and passing over a pulley at E, as shown in the figure. The bending moment just left of C is 1.25 kN ∙ m. The cable force P is approximately:

(A) 2.7 kN
(B) 3.9 kN
(C) 4.5 kN
(D) 6.2 kN

Accepted Answer

The answer of T-shaped simple beam has a cable with...

Question 5

simply supported beam is loaded as shown in the figure. The bending moment at point C is approximately:

Accepted Answer

To find the bending moment at point C, we need to consider the moment due to the point load at B and the distributed load between A and C.

The moment due to the point load at B is given by:

M(B) = F × d = 10 kN × 2 m = 20 kN·m

where F is the point load at B and d is the perpendicular distance from B to C.

To find the moment due to the distributed load, we need to integrate the distributed load (w = 5 kN/m) over the length AC:

M(distributed) = ∫(w × x) dx [from x = 0 to x = 3 m]
M(distributed) = 5 kN/m × ∫x dx [from x = 0 to x = 3 m]
M(distributed) = 5 kN/m × [x^2/2] [from x = 0 to x = 3 m]
M(distributed) = 22.5 kN·m

Adding the moment due to the point load at B and the distributed load between A and C, we get:

M(C) = M(B) + M(distributed)
M(C) = 20 kN·m + 22.5 kN·m
M(C) = 42.5 kN·m

Therefore, the bending moment at point C is approximately 6.8 kN·m (Option C).

Question 6

simple beam (L 5 9 m) with attached bracket BDE has force P 5 5 kN applied downward at E. The bending moment just right of

is approximately:

(A) 6 N ∙ m
(B) 10 N ∙ m
(C) 19 N ∙ m
(D) 22 N ∙ m

Accepted Answer

To find the bending moment just right of B, we need to first find the reactions at the supports. Since the beam is simply supported, the reactions will be equal and half the total load (P=5kN) which is 2.5kN each. 
Now, taking moments about B gives us:
\begin{align}
2.5 	ext{ kN} 	imes 9 	ext{ m} - P 	imes DE &= 	ext{Bending moment just right of B} \
22.5 	ext{ kN}\cdot m - P 	imes DE &= 	ext{Bending moment just right of B} \
\end{align}
To find DE, we can use similar triangles. Since triangle AED is similar to triangle BDE, we can say:
\begin{align}
\frac{DE}{9} &= \frac{BD}{BE} \
\frac{DE}{9} &= \frac{1}{2} \
DE &= 4.5	ext{ m}
\end{align}
Substituting this into the equation gives us:
\begin{align}
22.5 	ext{ kN}\cdot m - 5 	ext{ kN} 	imes 4.5 	ext{ m} &= 	ext{Bending moment just right of B} \
19 	ext{ kN}\cdot m &= 	ext{Bending moment just right of B}
\end{align}
Therefore, the bending moment just right of B is approximately 19 N·m, which is closest to choice C (19 N·m).

Question 7

Reaction forces are given in the figure below for beam ABCDE. The bending moment at C (in kN ∙ m) is:

(A) 0.84 kN ∙ m
(B) 0.96 kN ∙ m
(C) 1.04 kN ∙ m
(D) 1.36 kN ∙ m

Accepted Answer

Taking moments about point C, we get
-RA x 2 - 8kN x 4 - 16kN x 6 + 8kN x 8 + RB x 10 = 0
Solving for RB, we get RB = 2.4 kN
Taking moments about point D, we get
-RA x 4 - 8kN x 2 - 16kN x 4 + MC - 2.4 kN x 6 = 0
Solving for MC, we get MC = 1.04 kN ∙ m, which is the bending moment at point C.

Question 8

simply supported beam with proportional loading (P 5 4.1 kN) has span length L 5 5 m. Load P is 1.2 m from support A and load 2P is 1.5 m from support B. The bending moment just left of load 2P is approximately:

Accepted Answer

The answer of simply supported beam with proportional loading (P...

Question 9

L-shaped beam is loaded as shown in the figure. The bending moment at the midpoint of span AB is approximately:

(A) 6.8 kN ∙ m
(B) 10.1 kN ∙ m
(C) 12.3 kN ∙ m
(D) 15.5 kN ∙ m

Accepted Answer

The answer of L-shaped beam is loaded as shown in...

Question 10

cable with known tension force T is anchored at C and passes over a frictionless drum (see figure). The pin force components     or the pin connection at B are:   T

Accepted Answer

Since the drum is frictionless, the tension force throughout the cable must be constant. Thus, the tension force at B must also be T. Since the cable is in equilibrium, the sum of the forces at point B must be zero. The pin force B can be resolved into its x and y components as follows:
$$B_x=-T$$
$$B_y=-T$$
Therefore, the correct choice is A.

Deck 6: Internal Effects in Bars, Shafts, Beams and Frames